Biology for VCE Units 3 and 4 (Full Colour Interim) Student Edition

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STUDENT EDITION

BIOLOGY FOR VCE UNITS 3&4


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STUDENT WORKBOOK


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Meet the Writing Team

Tracey

Senior author

Lissa

Tracey Greenwood I have been writing resources for students since 1993. I have a Ph.D in biology, specialising in lake ecology and I have taught both graduate and undergraduate biology. Lissa Bainbridge-Smith I worked in industry in a research and development capacity for 8 years before joining BIOZONE in 2006. I have an M.Sc from Waikato University.

Author

Kent

Kent Pryor I have a BSc from Massey University majoring in zoology and ecology and taught secondary school biology and chemistry for 9 years before joining BIOZONE as an author in 2009.

The platypus (Ornithorhynchus anatinus) is endemic to Australia's eastern states and Tasmania. Human activity (damming, trapping, pollution, and fishing) resulted in a drop in platypus numbers, but conservation efforts have helped to restore their numbers. The platypus has a body temperature of 32°C, lower than most mammals, which is believed to be an adaptation to conserve energy when swimming in cold water. PHOTO: Joel Sartore (www.gettyimages.co.nz)

Author

Richard

Cover Photograph

Richard Allan I have had 11 years experience teaching senior secondary school biology. I have a Masters degree in biology and founded BIOZONE in the 1980s after developing resources for my own students.

Founder & CEO

First edition 2020 Full colour interim edition

ISBN 978-1-98-856647-4

Copyright Š2020 Richard Allan Published by BIOZONE International Ltd

Thanks to:

The staff at BIOZONE, including Clare Mansfield for design, Felix Hicks for illustration support, Paolo Curray and James Leggett for IT support, Anu Chauhan for logistics, Allan Young for office handling, and the BIOZONE sales team.

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BIOZONE Learning Media Australia

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electrical, mechanical, photocopying, recording or otherwise, without the permission of BIOZONE International Ltd. This workbook may not be re-sold. The conditions of sale specifically prohibit the photocopying of exercises, worksheets, and diagrams from this workbook for any reason.

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Purchases of this workbook may be made direct from the publisher:


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Contents

Using This Workbook ............................................. vi

Regulating biochemical pathways

Using the Tab System............................................. viii

Key knowledge .................................................. 53

Using BIOZONE's Website.......................................xi

34 Enzymes ........................................................... 54 35 Models of Enzyme Activity................................. 55

Answering Exam Questions......................................x

36 How Enzymes Work........................................... 56 37 Enzyme Kinetics ............................................... 57

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UNIT 3, AREA OF STUDY 1

Key knowledge.................................................... 1 1 2

Phospholipids and the Properties of Membranes. 2

3 4

How Do We Know? Membrane Structure............ 5

5 6

Membranes and the Export of Proteins............... 7

7

KEY TERMS AND IDEAS: Did You Get it? ......... 10

The Structure of the Plasma Membrane.............. 3 Investigating Transport Across Membranes......... 6 Endocytosis .......................................................... 9

Nucleic acids

Key knowledge................................................... 11

8 9

Amino Acids Make up Proteins .......................... 13 Separating Amino Acids by Chromatography..... 14

10 Protein Structure is Hierarchical........................ 15 11 Protein Shape is related to Function ................. 16 12 Comparing Globular and Fibrous Proteins......... 17 13 Protein Functional Diversity............................... 18 14 How Are Proteins Modified? ............................. 20 15 How Do We Know? DNA Carries the Code ...... 21 16 Nucleotides........................................................ 23 17 Nucleic Acids ..................................................... 24 18 Creating a DNA Model ....................................... 26 19 How Does DNA Replicate? ............................... 30 20 Modelling DNA Replication ............................... 32 21 The Evidence for DNA Structure........................ 35 22 What is Gene Expression? ............................... 36 23 The Genetic Code.............................................. 37 24 Cracking the Genetic Code................................. 38 25 Transcription in Eukaryotes................................. 39 26 mRNA Processing in Eukaryotes ....................... 41 27 Translation.......................................................... 42 28 Gene Expression Summary .............................. 44 29 KEY TERMS AND IDEAS: Did You Get it?......... 45

Gene structure and regulation Key knowledge ................................................. 46 30 Structural and Regulatory Genes ..................... 47 31 Eukaryotic Gene Structure and Regulation....... 48 32 Models of Gene Regulation in Prokaryotes....... 50 33 KEY TERMS AND IDEAS: Did You Get it? ........ 52

38 Investigating Catalase Activity .......................... 59 39 Enzyme Inhibition................................................ 60 40 Coenzymes ........................................................ 61 41 Achieving Metabolic Efficiency........................... 62 42 KEY TERMS AND IDEAS: Did You Get it?......... 63

Photosynthesis

Key knowledge ................................................. 64 43 The Role of Photosynthesis............................... 65 44 Chloroplasts....................................................... 66 45 The Bacterial Origin of Chloroplasts ................. 67 46 Pigments and Light Absorption.......................... 68 47 Separation of Pigments By Chromatography..... 69 48 Photosynthesis: Inputs and Outputs.................. 70 49 How Do We Know? Photosynthesis................... 72 50 Investigating Enzymes in Photosynthesis ......... 73 51 Factors Affecting Photosynthesis....................... 74 52 Overcoming Limiting Factors in Photosynthesis . 75 53 Investigating Photosynthetic Rate ..................... 76 54 KEY TERMS AND IDEAS: Did You Get it? ........ 77

Cellular respiration

Key knowledge ................................................. 78 55 The Role of ATP in Cells ................................... 79 56 ATP and Energy................................................. 80 57 Mitochondrial Structure and Bacterial Origin .... 81 58 Cellular Respiration: Inputs and Outputs............ 82 59 Measuring Respiration....................................... 84 60 Investigating Aerobic Respiration in Yeast ........ 85 61 Anaerobic Pathways.......................................... 86 62 Investigating Fermentation in Yeast ................... 87 63 KEY TERMS AND IDEAS: Did You Get it? ........ 89 64 Review: Unit 3, Area of Study 1 ........................ 90 65 Synoptic Assessment: Unit 3, Area of Study 1... 92

UNIT 3, AREA OF STUDY 2 Cellular signals

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Plasma membranes

Key knowledge ................................................. 94 66 What is Cell Signalling? .................................... 95 67 The Nature of Signalling Molecules................... 97 68 What is Signal Transduction? ............................ 98 69 Types of Signal Transduction ............................ 99

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70 Apoptosis: Programmed Cell Death ................ 101


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Contents

71 Malfunctions of Apoptosis................................ 103 72 KEY TERMS AND IDEAS: Did You Get it?....... 104

111 Types of Natural Selection .............................. 160 112 Selection for Human Birth Weight.................... 161 113 Selection for Skin Colour in Humans .............. 162 114 Directional Selection in Darwin's Finches ...... 164

Responding to antigens Key knowledge ............................................... 105

117 Evolution in Rock Pocket Mice......................... 167 118 Heterozygous Advantage ................................ 169

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73 The Nature of Antigens ................................... 106 74 Cellular and Non-Cellular Pathogens............... 108

115 Disruptive Selection in Darwin's Finches ........ 165 116 Directional Selection in Moths.......................... 166

75 The Body's Defences: An Overview ................ 109 76 Chemical Defences in Plant and Animals ....... 110 77 The Innate Immune Response......................... 112 78 Phagocytes and Phagocytosis......................... 114 79 Processing Antigens........................................ 115 80 The Lymphatic System ................................... 116 81 The Adaptive Immune System......................... 117 82 Clonal Selection .............................................. 119 83 Antibodies ....................................................... 120 84 KEY TERMS AND IDEAS: Did You Get it? ...... 121

Immunity

119 The Evolution of Antibiotic Resistance............. 170 120 Insecticide Resistance..................................... 171 121 Antigenic Variability in Pathogens.................... 172 122 The Founder Effect.......................................... 173 123 Genetic Bottlenecks......................................... 175 124 Genetic Drift .................................................... 176 125 Reproductive Isolation .................................... 177 126 Allopatric Speciation........................................ 179 127 Speciation in Australia ..................................... 181 128 Stages in Species Formation .......................... 182

Key knowledge ............................................... 122

129 Selective Breeding in Animals.......................... 183 130 Selection in Livestock...................................... 184

85 Acquired Immunity........................................... 123 86 Vaccines and Vaccination................................ 125

131 Selection and Population Change.................... 185 132 Selective Breeding in Crop Plants .................. 186

87 Vaccines Can Eliminate Infectious Disease .... 127 88 Autoimmune Disease: Multiple Sclerosis......... 128

133 Breeding Modern Wheat ................................. 188 134 KEY TERMS AND IDEAS: Did You Get it?....... 189

93 KEY TERMS AND IDEAS: Did You Get it? ...... 135 94 Review: Unit 3, Area of Study 2 ...................... 136 95 Synoptic Assessment: Unit 3, Area of Study 2. 137

UNIT 4, AREA OF STUDY 1 Changes in the genetic makeup of a population Key knowledge ............................................... 140 96 Sources of Variation ........................................ 141 97 Processes in Gene Pools ................................ 143 98 What is a Gene Mutation? .............................. 145 99 Reading Frame Shifts....................................... 146 100 Chromosome Mutations................................... 147 101 Gene Duplication as a Source of New Genes.. 148 102 Aneuploidy........................................................ 149 103 Polyploidy.......................................................... 150 104 Mutation in Populations: Sickle Cell ................ 151 105 Mutation in Populations: Cystic Fibrosis.......... 152 106 Beneficial Mutations in Humans ...................... 153 107 Mechanism of Natural Selection ..................... 154 108 Adaptation and Fitness.................................... 156 109 Gene Pool Exercise......................................... 157 110 Changes in a Gene Pool.................................. 159 Activity is marked:

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Changes in biodiversity over time

Key knowledge ............................................... 190 135 Evidence for Evolution ..................................... 191 136 Descent and Common Ancestry..................... 192

137 Changes in the Earth's Biodiversity Over Time........................................................ 194 138 Fossil Formation.............................................. 196 139 Relative Dating and the Fossil Record............. 198 140 Absolute Dating ............................................... 200 141 Transitional Fossils .......................................... 201 142 The Evolution of Whales.................................. 202 143 The Evolution of Horses................................... 203 144 Continental Drift and Evolution ....................... 204 145 Speciation and Dispersal of the Camelidae..... 208 146 Bigeographical Evidence................................. 209 147 Oceanic Island Colonisers............................... 210 148 Homologous Structures ................................... 212 149 Developmental Evidence for Evolution............. 213

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91 Monoclonal Antibodies .................................... 132 92 Herceptin: A Modern Monoclonal..................... 134

150 Patterns of Evolution........................................ 214 151 The Rate of Evolutionary Change ................... 215 152 Divergent Evolution in Ratites ......................... 216 153 Adaptive Radiation in Mammals....................... 218 154 Convergent Evolution....................................... 220 155 Extinction......................................................... 222 156 KEY TERMS AND IDEAS: Did You Get it? ...... 224

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89 HIV/AIDS: An Immune Deficiency Disease ..... 129 90 Allergies and Hypersensitivity.......................... 131


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Contents

Determining relatedness between species Key knowledge ............................................... 225

157 Determining Relatedness Using Proteins ....... 226 158 Determining Relatedness by DNA

UNIT 4, AREA OF STUDY 2 DNA manipulation Key knowledge ............................................... 283 195 What is DNA Manipulation?............................. 284 196 Making Recombinant DNA............................... 285 197 New Tools: Gene Editing with CRISPR............ 287

160 The Molecular Clock Theory............................ 230

198 DNA Amplification Using PCR......................... 288

161 Using Mitochondrial DNA................................. 231

199 Gel Electrophoresis.......................................... 290

162 What is a Phylogenetic Tree? .......................... 232

200 Interpreting Electrophoresis Gels.................... 291

163 Constructing Phylogenies Using Cladistics...... 233

201 Using Recombinant Plasmids In Industry........ 292

164 Constructing a Cladogram................................ 235

202 Using Recombinant Plasmids In Medicine....... 294

165 Master Genes and the Control

203 KEY TERMS AND IDEAS: Did You Get it?....... 296

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Hybridisation.................................................... 228

159 Genomic Comparisons and Relatedness ....... 229

of Development ............................................... 236

166 The Role of Master Genes............................... 237 167 KEY TERMS AND IDEAS: Did You Get it?....... 239

Biological knowledge and society

Key knowledge ............................................... 297

204 In Vivo Gene Cloning....................................... 298

Human change over time

205 DNA Profiling using PCR................................. 300

Key knowledge ............................................... 240

206 Forensic Applications of DNA Profiling............ 302

168 Primate Classification...................................... 241

207 Genetic Screening and Testing........................ 304

169 General Primate Characteristics...................... 242

208 What is Transgenesis?..................................... 306

170 The Primate Hand............................................. 244

209 The Applications of Transgenesis.................... 308

171 Hominoids and Hominins................................. 245

210 Engineering for Improved Nutrition.................. 309

172 Trends in Hominin Evolution: Overview ........... 246

211 Engineering for Insect Resistance................... 311

173 Trends in Skull Anatomy.................................. 248

212 Ethics of Genetic Modification.......................... 313

174 Trends in Brain Volume.................................... 250

213 Patterns of Disease.......................................... 315

175 Trends in Dentition........................................... 252

214 Containing the Spread of Disease................... 316

176 Bipedalism and Nakedness............................. 253

215 Emerging Disease............................................ 318

177 Adaptations for Bipedalism.............................. 254

216 Identifying and Treating Disease...................... 319

178 Cultural Evolution............................................. 256

217 Rational Drug Design....................................... 321

179 Trends in Palaeolithic Tool Cultures................. 257

218 Relenza: How an Antiviral Drug Works............ 323

180 Fire................................................................... 259

219 Antibiotics vs Antivirals.................................... 324

181 Shelter and Clothing........................................ 260

220 KEY TERMS AND IDEAS: Did You Get it?...... 326

182 Art and Spirituality............................................ 261

221 Review: Unit 4, Area of Study 2....................... 327

183 Mesolithic and Neolithic Cultures..................... 262

222 Synoptic Assessment: Unit 4, Area of Study 2. 328

184 Hominin Evolution: Probable Phylogenies....... 264 185 The Importance of Ardi.................................... 266

UNIT 4, AREA OF STUDY 3

186 New Findings: Denisovans............................... 268 187 New Interpretations: The Neanderthals ........... 269

Practical investigation

Key knowledge ............................................... 330

189 Problems with Dating: H. naledi....................... 271

223 Ethics and Issues of Research......................... 331

190 The Origin and Dispersal of Modern Humans. 272

224 Maintaining a Logbook..................................... 332

191 A Summary of Trends in Hominin Evolution..... 274

225 Analysis and Interpretation............................... 333

192 KEY TERMS AND IDEAS: Did You Get it?....... 278

226 Designing a Practical Investigation.................. 335

193 Review: Unit 4, Area of Study 1....................... 279

227 Presenting Your Findings.................................. 338

194 Synoptic Assessment: Unit 4, Area of Study 1. 281

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188 Problems With Interpretation: H. floresiensis... 270

Photo Credits and Acknowledgements .................. 340

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INDEX .................................................................... 341


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Using This Workbook

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This first edition of Biology for VCE Units 3 & 4 has been specifically written to meet the content and skills requirements of the Victorian Certificate of Education, Biology, Units 3 and 4. Each chapter introduction provides you with a concise guide to the knowledge and skills requirements for each area of key knowledge. Each key knowledge point is matched to the activity or activities addressing it. Activities complementing practical skills are identified in the chapter introductions by a code (PRAC) and are designed to provide background and familiarity with apparatus, techniques, experimental design, and interpretation of results. A wide range of activities will help you to build on what you already know, explore new topics, work collaboratively, and practise your skills in data handling and interpretation. We hope that you find the workbook valuable and that you make full use of its features.

The outline of the chapter structure below will help you to navigate through the material in each chapter.

Introduction

Activities

Literacy

Review

• A check list of key knowledge • A list of key terms

• The KEY IDEA provides your focus for the activity • Annotated diagrams help you understand the content • Questions review the content of the page

• Almost all chapters have a literacy activity based on the introductory key terms list • Several types of activities test your understanding of biological terms

• Create your own summary for review. • Hints help you to focus on what is important. • Your summary will consolidate your understanding of the content in the area of study

Synoptic Assessment

• Synoptic assessments conclude the unit and area of study covered in the workbook. • These enable you to practise your written exam skills

Linkages are made between ideas in separate activities

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Structure and organisation of chapters


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Understanding the activity coding system and making use of the online material identified will enable you to get the most out of this resource. The chapter content is structured to build knowledge and skills but this structure does not necessarily represent a strict order of treatment. Be guided by your teacher, who will assign activities as part of a wider programme of independent and group-based work.

Look out for these features and know how to use them: Free response questions allow you to use the information provided to answer questions about the content of the activity, either directly or by applying the same principles to a new situation. In some cases, an activity will assume understanding of prior content.

The activities form most of this workbook. They are numbered sequentially and each has a task code identifying the skill emphasised. Each activity has a short introduction with a key idea identifying the main message of the page. Most of the information is associated with pictures and diagrams, and your understanding of the content is reviewed through the questions. Some of the activities involve modelling and group work.

A TASK CODE on the page tab identifies the type of activity. For example, is it primarily information-based (KNOW), or does it involve modelling (PRAC) or a skill (SKILL)? A full list of codes is given on the following page but the codes themselves are relatively self explanatory.

WEB tabs at the bottom of the activity page alert the reader to the Weblinks resource, which provides external, online support material for the activity, usually in the form of an animation, video clip, photo library, or quiz. Bookmark the Weblinks page (see next page) and visit it frequently as you progress through the workbook.

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The chapter introduction provides you with a summary of the knowledge and skills requirements for the topic, phrased as a set of learning outcomes. Use the check boxes to identify and mark off the points as you complete them. The chapter introduction also provides you with a list of key terms for the chapter, from which you can construct your own glossary as you work through the activities.

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LINK tabs at the bottom of the activity page identify activities that are related in that they build on content or apply the same principles to a new situation.


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Using the Tab System

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The tab system is a useful system for quickly identifying related content and online support. Links generally refer to activities that build on the information in the activity in depth or extent. A link may also reflect on material that has been covered earlier as a reminder for important terms that have already been defined. In the example below for the activity "Antibodies", the weblink 83 shows how antibodies are produced. Activity 78 defines some of the terms used on the page (in case students want to refresh their memory) and activity 81 covers the role of B cells in antibody production. The weblinks code is always the same as the activity number on which it is cited. On visiting the weblinks page (below), find the number and it will correspond to one or more external websites supporting the activity's content. Corrections and clarifications to current editions are also identified on the weblinks page.

Activities are coded

KNOW = content you need to know EXT

= extension material for accelerated students

SKILL = data handling and interpretation

PRAC = a paper practical or a practical focus REFER = reference - use this for information REVISE = review the material in the section TEST = test your understanding

Link

Weblinks

Bookmark the weblinks page: www.biozone.com.au/weblink/ VCE2-9414

Access the external URL for the activity by clicking the link

Connections are made between activities in different sections of the syllabus that are related through content or because they build on prior knowledge.

This WEBLINKS page provides links to external websites with supporting information for the activities. Mostly, these are animations and video clips directly relevant to some aspect of the activity on which they are cited. The weblinks page also provides access to annotated 3D models to explore specific aspects of your course. Bookmark this page as it is not accessible directly from BIOZONE's the website.

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Annotated 3D models organised under broad topics

Chapter in the workbook

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Hyperlink to the external website page.


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Using BIOZONE's Website Contact us with questions, feedback, ideas, and critical commentary. We welcome your input.

Use Google to search for websites of interest. The more precise your search words are, the better the list of results. Be specific, e.g. "biotechnology medicine DNA uses", rather than "biotechnology".

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Access the BIOLINKS database of web sites directly from the homepage of our website.

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Biolinks is organised into easy-to-use sub-sections relating to general areas of interest. It's a great way to quickly find out more on topics of interest.


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Answering Exam Questions The external exams require you to demonstrate your understanding of a particular concept by providing a written paragraph or essay. Open answer questions (meaning there is no definitive answer) are designed so that you can demonstrate your level of understanding. The question may give you some guidance as to what you should include in your answer, such as definitions of certain terms or advice to provide specific examples. In order to gain the highest possible mark in these questions, you need to lay out your answer in a clear and logical way so that the examiner can easily see how you have demonstrated your understanding of the topic.

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The difference between you obtaining an a low, mid, or high grade depends on how well you demonstrate your understanding of a concept.

• Defining, drawing, annotating, or giving a description demonstrates a basic understanding of the material. • Explaining how a process works, why it works, and how changes to it may affect an outcome shows a deeper understanding of how the system works in that situation. • Linking biological ideas, comparing and contrasting, analysing, or justifying shows both a deep understanding and an ability to translate that understanding to a new situation.

The following example question shows how an answer can be built up from a simple definition, through explanation, to comparisons and linking of ideas. Gene mutations may affect the protein formed during gene expression. Using an example, discuss the significance of mutations and the effect they may have on the fitness of an organism:

The term gene mutation is defined

A gene mutation is a change to the base pairs in a DNA sequence that makes up a gene.

Mutations can involve changes to a single nucleotide (called point mutations), or changes to a triplet by the insertion or deletion of a nucleotide. These are called frame shift mutations.

Example used to link the biological idea of how a beneficial mutation affects fitness

Mutations can be harmful, beneficial, or have no apparent effect. The type of mutation can affect the fitness of an organism in its current environment. Fitness is a measure of an organism's ability to survive and reproduce (contribute genes to the next generation).

An example of a beneficial mutation improving fitness is antibiotic resistance in bacteria. Some gene mutations in bacteria confer resistance to specific antibiotics. Bacteria with the beneficial gene mutation are not killed by the antibiotic and survive to reproduce. When they reproduce, the gene is passed on to the next generation and resistance to that particular antibiotic spreads quickly through the population. In contrast, bacteria without the gene mutation have decreased fitness for that particular antibiotic and are killed by it (so can not reproduce).

The term fitness is defined

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Effect of mutations on fitness is explained.

Mutations can affect the protein formed, but some types of gene mutations are more likely to significantly alter proteins than others. Frame shift mutations can cause significant changes because the instructions for making a protein can be changed, producing a completely different amino acid sequence or a protein that is partly completed. Point mutations may also result in a different amino acid being inserted, but sometimes the mutation still codes for the same amino acid and the protein produced is the same.

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The significance of different types of gene mutations is discussed.

Different types of gene mutations are described.

Comparing and contrasting the effect of a mutation on a population's fitness


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Plasma membranes

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Unit 3 Outcome 1

Key terms

The fluid mosaic model

active transport

Key knowledge

carrier protein

channel protein

c

1

Describe the fluid mosaic model of membrane structure. Include reference to the significance of phospholipid orientation and the role of transmembrane proteins in the integral structure of the membrane.

1 2 3

c

2

Recall that the internal membranes of cells, e.g. the membranes of membranous organelles, have the same basic structure as the plasma membrane.

5

c

3

Recall the principles of membrane transport by passive and active processes. Explain the movement of hydrophilic and hydrophobic substances across the plasma membrane, including reference to molecule size and polarity.

2

c

4

PRAC

concentration gradient diffusion

endocytosis exocytosis

facilitated diffusion Golgi apparatus

Activity number

hydrophilic

Analyse experimental data comparing the rates of movement of substances of different size and polarity across plasma membranes.

4

hydrophobic hypertonic hypotonic ion

ion pump isotonic

non-polar

osmolarity

Mnolf

osmosis

The export of proteins across membranes

partially permeable (= selectively-permeable)

Key knowledge

passive transport

c

5

Explain how proteins destined for secretion from the cell are packaged and exported across the cell's plasma membrane by exocytosis. Include an explanation of the role of ribosomes, rough endoplasmic reticulum (rER), and the Golgi apparatus and its associated vesicles.

5

c

6

Recognise exocytosis as a type of active transport and explain what this means.

5

plasma membrane polar

Endocytosis

Activity number

Key knowledge

7

Explain what is meant by endocytosis and distinguish it from exocytosis. Recognise endocytosis as a type of active transport and explain what this means.

6

c

8

Distinguish between endocytosis of fluids (pinocytosis) and solids (phagocytosis). Describe examples of when substances are moved across the plasma membrane by these mechanisms.

6

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phospholipid

Activity number


2

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Phospholipids and the Properties of Membranes

Key Idea: Phospholipids are important components of cellular membranes. They are made up of a hydrophobic head region and a hydrophilic tail region, making them amphipathic. Phospholipids consist of a glycerol attached to two fatty acid chains and a phosphate (PO43-) group. Phospholipids

naturally form bilayers in aqueous solutions and are the main component of cellular membranes. The fatty acid tails can be saturated (forming straight chains) or unsaturated (kinked chains). The level of phospholipids with saturated or unsaturated tails affects the fluidity of the phospholipid bilayer.

Phospholipids and membranes

The phosphate end of the phospholipid is attracted to water (it is hydrophilic) while the fatty acid end is repelled (hydrophobic). In an aqueous environment, the hydrophobic ends turn inwards in the membrane to form a bilayer. Fatty acids containing double C=C bonds are unsaturated. This causes a "kink" in the chain.

Phospholipids are amphipathic (have hydrophobic and hydrophilic regions). This means that they will spontaneously form bilayers when in aqueous environments and so form the outer boundary of cells and organelles. Modifications to the hydrophobic ends of the phospholipids regulate the fluidity of the bilayer. The greater the number of double bonds in the hydrophobic tails, the greater the fluidity of the membrane.

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Phospholipids

Phosphate group

+

Hydrophilic head

CH2

N (CH3)3

CH2 O

P

O

–

O

O

CH

O

O

C

O C

Hydrophobic tails

H2C

CH2

Membrane containing only phospholipids with saturated fatty acid tails.

O

Double bond

Membrane containing phospholipids with unsaturated fatty acid tails. The fact that the phospholipids do not stack neatly together produces a more fluid membrane that may remain fluid even at low temperatures.

Fatty acid tail

1. (a) How do the properties of phospholipids contribute to their role in forming the structural framework of membranes?

(b) Explain why phospholipid bilayers containing many phospholipids with unsaturated tails are particularly fluid:

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WEB

KNOW

1

LINK

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2. Suggest how the cell membrane structure of an Arctic fish might differ from that of tropical fish species:

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3

The Structure of the Plasma Membrane

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2

Key Idea: A cellular membrane is made of a phospholipid bilayer with proteins of different sorts embedded in it. The cell surface (or plasma) membrane encloses the

cell's contents and regulates many of the cell's activities. Importantly, it controls what enters and leaves the cell by the use of carrier and channel proteins.

Simple membrane structure

Lipid soluble molecules, e.g. gases and steroids, can move through the membrane by diffusion, down their concentration gradient.

Carrier proteins permit the passage of specific molecules by facilitated diffusion or active transport.

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CO2

Phospholipids naturally form a bilayer.

Phosphate head is hydrophilic

Fatty acid tail is hydrophobic

H2O

Glucose

Na+

Channel proteins form a pore through the hydrophobic interior of the membrane to enable water soluble molecules to pass by facilitated diffusion.

Cholesterol molecule maintains membrane integrity, preventing it becoming too fluid or too firm.

Water molecules pass between the phospholipid molecules by osmosis.

What can cross a lipid bilayer?

Gases

Hydrophobic molecules

CO2

Benzene

Small polar molecules

Large polar molecules

Charged molecules

Na+

H2O

Glucose

O2

Cl-

Ca2+

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Lipid soluble molecules diffuse into and out of the membrane unimpeded.

Polar molecules are small enough to diffuse through. Aquaporins increase rate of water movement.

Cannot directly cross the membrane. Transport (by facilitated diffusion or active transport) involves membrane proteins. LINK

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Small uncharged molecules can diffuse easily through the membrane.

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Ethanol

4

Ions can be transported across the membrane by ion channels (passive) or ion pumps (active).

LINK

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What do proteins in the cell surface membrane really look like?

Extracellular

The receptor (darker) is bound to intracellular G protein (lighter).

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Vossman cc 3.0

A2-33 cc 3.0

The structure of membrane proteins enables them to perform their particular function in transport, cell signalling, or cell recognition. The proteins are integral to the membrane, and often have parts of their structure projecting from both internal and external sides of the membrane.

Aquaporins are a special type of channel protein that speed up the passage of water molecules across the membrane. Their tertiary structure creates a pore through the centre of the protein through which molecules can pass.

The GLUT1 glucose transporter is a carrier protein that facilitates the transport of glucose across the plasma membranes of mammalian cells. It increases the rate of glucose transport by 50,000X (high enough to supply the cell's energy needs).

Intracellular

G-protein coupled receptors are proteins involved in signalling pathways. A signal molecule binds to the receptor protein outside the cell to trigger a reaction involving intracellular G protein. The receptor in this example binds to adrenaline.

Cell surface antigens provide an identifiable cell signature so that the body can distinguish between its own cells and foreign molecules. They are often glycoproteins. The image above shows how the antigens project from the membrane.

1. What is the purpose of carrier proteins in the membrane?

2. What is the purpose of channel proteins in the membrane?

3. Identify the molecule(s) that:

(a) Can diffuse through the plasma membrane on their own:

(b) Can diffuse through the membrane via channel proteins:

(c) Must be transported across the membrane by carrier proteins:

(a) Aquaporins:

(b) GLUT1 protein:

(c) G protein:

(d) Cell surface antigens:

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4. Describe the role of the following proteins in the plasma membrane:

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How Do We Know? Membrane Structure

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Key Idea: The freeze-fracture technique for preparing and viewing cellular membranes has provided evidence to support the fluid mosaic model of the plasma membrane.

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Cellular membranes play many extremely important roles in cells and understanding their structure is central to understanding cellular function. Moreover, understanding the structure and function of membrane proteins is essential to understanding cellular transport processes, and cell recognition and signalling. Cellular membranes are far too small to be seen clearly using light microscopy, and certainly any detail is impossible to resolve. Since early last century, scientists have known that membranes were composed of a lipid bilayer with associated proteins. The original model of membrane structure, proposed by Davson and Danielli, was the unit membrane (a lipid bilayer coated with protein). This model was later modified by Singer and Nicolson after the discovery that the protein molecules were embedded within the bilayer rather than coating the outside. But how did they find out just how these Cleaving the membrane molecules were organised? The answers were provided with electron microscopy, and one technique in particular – freeze fracture. As the name implies, freeze fracture, at its very simplest level, is the freezing of a cell and then fracturing it so the inner surface of the membrane can be seen using electron microscopy. Membranes are composed of two layers of phospholipids held together by weak intermolecular bonds. These split apart during fracture. The procedure involves several steps:

Cells are immersed in chemicals that alter the strength of the internal and external regions of the plasma membrane and immobilise any mobile macromolecules.

The cells are passed through a series of glycerol solutions of increasing concentration. This protects the cells from bursting when they are frozen. The cells are mounted on gold supports and frozen using liquid propane.

Razor blade

-150°C. A

The cells are fractured in a helium-vented vacuum at razor blade cooled to -170° C acts as both a cold trap for water and the fracturing instrument.

Proteins leave bumps and holes in the membrane when it is cleaved

The surface of the fractured cells may be evaporated a little to produce some relief on the surface (known as etching) so that a three-dimensional effect occurs. For viewing under an electron microscope (EM), a replica of the cells is made by coating them with gold or platinum to ~3 nm thick. A layer of carbon around 30 nm thick is used to provide contrast and stability for the replica. The samples are then raised to room temperature and placed into distilled water or digestive enzymes, which separates the replica from the sample. The replica is then rinsed in distilled water before it is ready for viewing.

The freeze fracture technique provided the necessary supporting evidence for the current fluid mosaic model of membrane structure. When cleaved, proteins in the membrane left impressions that showed they were embedded into the membrane and not a continuous layer on the outside as earlier models proposed.

50 nm

Photo: Louisa Howard and Chuck Daghlian, Dartmouth College

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1. Explain how freeze-fracture studies provided evidence for our current model of membrane structure:

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2. The Davson and Danielli model of membrane structure was the unit membrane; a phospholipid bilayer with a protein coat. Explain how the freeze-fracture studies showed this model to be flawed:

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Investigating Transport Across Membranes

Key Idea: The rate of diffusion of molecules through the plasma membrane can be determined by measuring the change in light absorbance as a solution of red blood cells haemolyses. How a cell behaves when suspended in a solution depends on whether or not the molecules or ions in the solution can cross the plasma membrane. If red blood cells (RBCs) are suspended in a concentrated solution of molecules that can

readily diffuse across the membrane, the molecules will enter the RBCs by diffusing down their concentration gradient. This will draw water into the RBCs (by osmosis) and they will burst (haemolyse). When the RBCs burst, the cellular material settles out of suspension and the solution becomes clear. By using a spectrophotometer to measure the rate at which the solution becomes clear, it is possible to determine the rate at which the molecules are crossing the plasma membrane.

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The aim

0.3 mol L-1 solution

To investigate how the size and membrane solubility of molecules affects the rate of diffusion across the plasma membrane.

The method

• 0.3 mol L-1 solutions of glucose, sucrose, urea, and glycerol were prepared (this concentration is greater than the cell internal concentration). A blank solution of distilled water was also prepared. Molecular weights (MW) are as follows: glucose (MW 180), sucrose (MW 342), urea (MW 60), and glycerol (MW 92). • Both urea and glycerol readily diffuse across the plasma membrane. Glucose is transported across the membrane by a carrier protein. • 3 mL of each solution was mixed with 0.1 mL of a sheep RBC suspension and added to cuvettes. The cuvettes were placed into a spectrophotometer and absorbance measured over 15 minutes. The results are plotted below:

Diffusing molecule Water molecule

Hypertonic solution

Absorbance vs time for sheep RBCs

0.7 0.6

Membrane-penetrating molecule crosses membrane, increasing the cell's internal solute concentration and drawing water in by osmosis.

Absorbance

0.5 0.4

Glycerol Glucose Sucrose Urea

0.3

Cell bursting

0.2 0.1

Linda Scott et al 1993 Dept. Bio. Hartwick College, NY.

0

0

5 10 20 30 60 120

240

320

480

600

720

840

Time (seconds)

Increased cellular volume causes the cell to burst (haemolysis). The cellular material settles out.

1. (a) Which molecule crosses the membrane the fastest?

(b) Which molecule appears to be unable to cross the plasma membrane?

(c) List the molecules in order of their ability to cross the plasma membrane (fastest to slowest):

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2. (a) What is the largest molecule used in the experiment?

(b) What is the smallest molecule used in the experiment?

(c) How does size affect the rate at which molecules can cross the plasma membrane?

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3. Why don't the RBCs in the glucose solution haemolyse even though glucose is transported across the membrane?

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Membranes and the Export of Proteins

Key Idea: The synthesis, packaging and movement of macromolecules inside the cell involves coordination between several membrane-bound organelles. Many proteins need to be modified in order to become

As it enters the cisternal space inside the ER, it folds up into its correct 3-dimensional shape.

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3

functional. This modification takes place in the rough endoplasmic reticulum (rER). From the rER, proteins are transported to the Golgi where the protein is further modified before packaged and shipped to its final destination.

Ribosomes on the surface of the endoplasmic reticulum (ER) translate mRNA into a polypeptide chain.

1

4

Ribosome

2

The chain is threaded through the ER membrane into the cisternal space, possibly through a pore.

6

These vesicles are received by the Golgi apparatus which further modifies, processes, and packages the proteins. Proteins move through the Golgi stack from one side of the organelle to the other, undergoing modification by different enzymes along the way. They are eventually shipped to the cell's surface, where they can be exported from the cell by exocytosis.

5

Transport vesicle

Most proteins destined for secretion are glycoproteins (i.e. they are proteins with carbohydrates added to them). Attachment of carbohydrate to the protein by enzymes (glycolysation) occurs in the rER.

Proteins destined for secretion leave the ER wrapped in transport vesicles which bud off from the outer region of the ER.

Transport vesicle

(a) Ribosomes:

(b) Endoplasmic reticulum:

(c) Transport vesicles:

(d) Golgi apparatus:

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1. Explain the role of each of the following organelles in the production and transport of proteins:

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Exocytosis

Exocytosis (below) is an active transport process in which a secretory vesicle fuses with the plasma membrane and expels its contents into the extracellular space. In multicellular organisms, various types of cells (e.g. endocrine cells and nerve cells) are specialised to manufacture products, such as proteins, and then export them from the cell to elsewhere in the body or outside it.

3

Plasma membrane

The contents of the vesicle are expelled into the extracellular space.

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Exocytosis (and its counterpart endocytosis) require energy because they involve movement of cytoskeletal proteins.

2

1

From Golgi apparatus

Vesicle fuses with the plasma membrane.

Vesicle from the Golgi carrying molecules for export moves to the perimeter of the cell.

Nerve cell

NT

Nerve cell

Golgi apparatus forming vesicles

The transport of Golgi vesicles to the edge of the cell and their expulsion from the cell occurs through the activity of the cytoskeleton. This requires energy (ATP).

Exocytosis is important in the transport of neurotransmitters into the junction (synapse) between nerve cells to transmit nervous signals.

Fungi and bacteria use exocytosis to secrete digestive enzymes, which break down substances extracellularly so that nutrients can be absorbed (by endocytosis).

2. (a) What is the purpose of exocytosis?

(b) How does it occur?

(b)

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(a)

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3. Describe two examples of the purpose of exocytosis in cells:

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Endocytosis

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Key Idea: Endocytosis is an active transport process in which the cell engulfs material and draws it in. Endocytosis is a type of active transport in which the plasma

membrane folds around a substance to transport it across the plasma membrane into the cell. The ability of cells to do this is a function of the flexibility of the plasma membrane.

Material (solids or fluids) that are to be brought into the cell are engulfed by an infolding of the plasma membrane.

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Plasma membrane

The vesicle carries molecules into the cell. The contents may then be digested by enzymes delivered to the vacuole by lysosomes.

Vesicle buds inwards from the plasma membrane

Phagocytosis (or ‘cell-eating’) involves the cell engulfing solid material to form large phagosomes or vacuoles (e.g. food vacuoles). It may be non-specific or receptormediated. Examples: Feeding in Amoeba, phagocytosis of foreign material and cell debris by neutrophils and macrophages.

HIV particle

Receptor mediated endocytosis is triggered when certain metabolites, hormones, or viral particles bind to specific receptor proteins on the membrane so that the material can be engulfed. Examples: The uptake of lipoproteins by mammalian cells and endocytosis of viruses (above).

Dartmouth College

CDC

Dartmouth College

Receptors and pit beginning to form

Pinocytosis (or ‘cell-drinking’) involves the non-specific uptake of liquids or fine suspensions into the cell to form small pinocytic vesicles. Pinocytosis is used primarily for absorbing extracellular fluid. Examples: Uptake in many protozoa, some cells of the liver, and some plant cells.

1. What is the purpose on endocytosis?

2. Is endocytosis active or passive transport?

(a) Phagocytosis:

(b) Receptor mediated endocytosis:

(c) Pinocytosis:

4. Explain how the plasma membrane can form a vesicle:

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3. Describe the following types of endocytosis:

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KEY TERMS AND IDEAS: Did You Get It?

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1.

Test your vocabulary by matching each term to its definition, as identified by its preceding letter code.

active transport carrier protein

channel protein

A A partially-permeable phospholipid bilayer forming the boundary of all cells. B

The movement of substances across a biological membrane without energy expenditure.

C The passive movement of molecules from high to low concentration.

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D A transmembrane protein that moves ions across a plasma membrane against their concentration gradient.

diffusion

E Protein that forms a pore through the hydrophobic interior of the membrane to enable water soluble molecules to pass by facilitated diffusion.

endocytosis

F The energy-requiring movement of substances across a biological membrane against a concentration gradient.

ion pump osmosis

passive transport

plasma membrane

G Active transport in which molecules are engulfed by the plasma membrane, forming a phagosome or food vacuole within the cell.

H Passive movement of water molecules across a partially permeable membrane down a concentration gradient.

I

Protein that permits the passage of specific molecules by facilitated diffusion or active transport.

2. The diagram (right) symbolically represents a phospholipid.

(a) Label the hydrophobic and hydrophilic regions of the phospholipid, and the phosphate group and fatty acids:

(b) Explain how the properties of the phospholipid molecule result in the bilayer structure of membranes:

3. Match the statements in the table below to form a complete paragraph: Transport of molecules though the plasma membrane...

....to the movement of molecules or ions against their concentration gradient.

Active transport requires the input of energy...

...high concentration to low concentration (down a concentration gradient).

Passive transport involves the movement of molecules from... Simple diffusion can occur...

Facilitated diffusion involves proteins in the plasma membrane...

Active transport involves membrane proteins, which couple the energy provided by ATP...

...can be active or passive.

...directly across the membrane.

...which help molecules or ions to move through. ...whereas passive transport does not.

TEST

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4. Describe the passage of a protein from its translation to its secretion from a cell by exocytosis:

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Nucleic acids

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Unit 3 Outcome 1

Key terms

Nucleic acids code for proteins

adenine

Key knowledge

anticodon

base-pairing rule chromatography coding strand

c

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Amino acids are the structural units that make up proteins. They have different properties and these determine how they interact to form functional proteins.

8

c

2

Describe how nucleic acids encode the instructions for the synthesis of proteins in cells, including reference to the relationship between the base sequence in a nucleic acid and the order of the amino acids in a polypeptide chain.

8

c

3

PRAC

Use chromatography to separate amino acids based on their properties.

9

cytosine

denaturation

Activity number

DNA

double-helix

Proteins carry out diverse roles

exons

Key knowledge

fibrous protein gene

gene expression

c

4

Explain what is meant by the proteome and distinguish it from the genome.

c

5

Using examples, describe the functional diversity of proteins. Describe examples of how proteins have roles in structure, catalysis, regulation, movement, defence, and transport.

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Describe how proteins may be modified in order to achieve their final function, e.g. by cleavage, or by the addition of carbohydrate, phosphate, or lipid groups.

genetic code genome

Activity number

globular protein

13

11 12 13

14

guanine

hydrogen bonding introns

nucleic acids nucleotides

peptide bond polypeptide

primary structure protein

proteome

Protein structure is hierarchical

purine

Key knowledge

pyrimidine

quaternary structure

Activity number

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Explain the functional importance of the four levels of protein structure: •

Primary (1°) structure: recall what determines the amino acid sequence.

RNA (mRNA, rRNA, tRNA)

Secondary (2°) structure: what is the role of hydrogen bonding in the 3D arrangement of segments of the polypeptide chain (e.g. the alpha helix)?

secondary structure

semi-conservative replication

Tertiary (3°) structure: what roles do disulfide and ionic bonds play in the final geometric shape a protein takes when it folds up?

Quaternary (4°) structure: the number and specific arrangement of multiple folded protein subunits. Not all proteins have a quaternary structure.

template strand thymine transcription translation uracil

c

8

Explain what is meant by denaturation and relate this to the protein's functional shape. Describes factors that can cause denaturation and describe examples.

11

c

9

Distinguish between globular and fibrous proteins as general structural categories. Relate the structure of a protein to its functional role in the cell (see #5). Give examples to illustrate how the arrangement of amino acids in the polypeptide chain contributes to a protein's function, e.g. in forming a hydrophilic pore in the membrane or the catalytic region (active site) of an enzyme.

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tertiary structure

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ribosome

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CL N AS OT SR F OO OR M US E Eva Santini, Giovanna Canu CC 4.0

Amino acids are joined by condensation Key knowledge

c

10

Explain how a polypeptide chain is synthesised from amino acid monomers by condensation polymerisation. Identify the bond formed by this process.

c

11

Know that polypeptides are broken down by hydrolysis.

DNA and RNA

Activity number 8 22 8 Activity number

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Key knowledge

15

c

12

Outline the experimental evidence that DNA was the genetic information (not proteins as was originally thought). Recognise that the understanding of DNA's role as the carrier of the code was a consequence of the successive investigations of a number of researchers.

c

13

Describe the structure of DNA and explain the evidence for it. In what way did the structure of DNA provide a mechanism for its self-replication?

c

14

Describe the Watson-Crick double-helix model of DNA. Include reference to the base-pairing rule, the anti-parallel strands, and the role of hydrogen bonding between purines and pyrimidines.

c

15

PRAC

c

16

Describe the structure of the three types of RNA (mRNA, tRNA, and rRNA) and their functional roles in cellular activities. Identify similarities and differences between RNA and DNA.

17

c

17

Describe the synthesis of nucleic acids by condensation polymerisation, including reference to the role of polymerase enzymes.

17

c

18

In more detail than #17, describe the semi-conservative replication of DNA, including the significance of synthesis in the 5’ to 3’ direction.

19

c

19

Use a model to reproduce the results of Meselson and Stahl's experiments that proved that DNA replication was semi-conservative.

20

EII

Create a model of DNA to demonstrate the base pairing rule.

The genetic code and gene expression Key knowledge

16 17 21 17

18

Activity number

20

Describe the features of the genetic code, including: • The 4-letter alphabet and the 3-letter triplet code (codon) of base sequences. • The non-overlapping, linear nature of the code, which is read from start to finish in one direction. The specific punctuation codons and their significance. • The universal nature and degeneracy of the code.

23

c

21

Explain how researchers determined that the genetic code was a triplet code.

24

c

22

Explain what is meant by a gene. Describe the steps involved in gene expression including transcription, RNA processing (eukaryotic cells), and translation. Identify where in the cell each of these steps occurs. Recognise transcription and translation as the two stages of gene expression. Describe the simple one gene-one polypeptide model for gene expression.

22 25 - 28

c

23

Outline what is involved in RNA processing, including reference to intron removal and exon splicing. How does alternative exon splicing account for the difference in the size of the proteome, relative to the number of identified genes in the human genome?

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8 Amino Acids Make up Proteins

Key Idea: Amino acids join together in a linear chain by condensation reactions to form polypeptides. The sequence of amino acids in a protein is defined by a gene and encoded in the genetic code. In the presence of water, they can be broken apart by hydrolysis into their constituent amino acids.

The structure and properties of amino acids

R

O

Carboxyl group

NH 2 CH2 CH2 CH2 CH22

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Chemically variable 'R' group

Amino acids are the basic units from which proteins are made. Twenty amino acids commonly occur in proteins and they can be linked in many different ways by peptide bonds to form a huge variety of polypeptides. Proteins are made up of one or more polypeptide molecules.

Amine group

NH2

Carbon atom

C

C

OH

H

SH

CH2

COOH CH 2

Hydrogen atom

All amino acids have a common structure (above), but the R group is different in each kind of amino acid (right). The property of the R group determines how the amino acid will interacts with others and how the amino acid chain will fold up into a functional protein. For example, the hydrophobic R groups of soluble proteins are folded into the protein's interior, while the hydrophilic groups are arranged on the outside.

Cysteine The 'R' group of cysteine forms disulfide bridges with other cysteines to create cross linkages in a polypeptide chain.

Lysine The 'R' group of lysine gives the amino acid an alkaline property.

Aspartic acid The 'R' group of aspartic acid gives the amino acid an acidic property.

Polypeptides are made by condensation and broken down by hydrolysis

Amino acids are linked by peptide bonds to form polypeptide chains of up to several thousand amino acids. Peptide bonds form between the carboxyl group of one amino acid and the amine group of another (right). Water is formed as a result of this bond formation so the reaction is called a condensation. The sequence of amino acids in a polypeptide is called the primary structure and is determined by the order of nucleotides in DNA and mRNA.

A Polypeptide Chain

Peptide bond

Peptide bond

Peptide bond

Peptide bond

Peptide bond

Two amino acids

R1

H2N

R2

O

H2N

C

C

OH

H

OH

H

Condensation Two amino acids are joined to form a dipeptide with the release of a water molecule.

H2N

O

C

C

Hydrolysis When a dipeptide is split, a water molecule provides a hydrogen and a hydroxyl group.

R1

O

H

R2

C

C

N

C

O

C

H

H

+ H2O

OH

Dipeptide

1. (a) What makes each of the amino acids in proteins unique? (b) What is the primary structure of a protein?

(c) What determines the primary structure?

(d) How do the sequence and composition of amino acids in a protein influence how a protein folds up?

(b) How is this bond formed?

(d) How are di- and polypeptides broken down?

(c) Circle this bond in the dipeptide above:

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2. (a) What type of bond joins neighbouring amino acids together?

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9 Separating Amino Acids by Chromatography

Key Idea: Paper chromatography is used to separate substances in a sample solution. Chromatography is a technique used to separate a mixture of molecules. Chromatography involves passing a mixture dissolved in a mobile phase (a solvent) through a

stationary phase, which separates the molecules according to their specific characteristics (e.g. size or charge). Paper chromatography is a simple technique in which porous paper serves as the stationary phase, and a solvent, either water or ethanol, serves as the mobile phase.

Paper chromatography

Using chromatography to separate amino acids Mixtures of amino acids can be separated by paper chromatography. The Rf values for amino acids separated using paper chromatography are given below:

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Set up and procedure

The chromatography paper is folded so it can be secured by the bung inside the test tube. The bung also prevents the solvent evaporating. Chromatography paper may be treated with chemicals to stain normally invisible pigments.

A spot of concentrated sample is added using a pipette and suspended above the solvent. As the solvent travels up the paper it will carry the sample with it. The distance the sample travels depends on its solubility.

Glycine: 0.50 Alanine: 0.70 Arginine: 0.72 Leucine: 0.91

A student was given a solution containing two unknown amino acids. They separated them by paper chromatography and obtained the results below. Amino acid A

Amino acid B

A pencil line is used to show the starting point.

Sample application point

Solvent

Determining Rf va lues

y = 5 cm

To identify the substances in a mixture an Rf value is calculated using the equation:

Rf =

y

x

2.5 cm

Distance travelled by the spot (x)

3.5 cm

Distance travelled by the solvent (y)

These Rf values can then be compared with Rf values from known samples or standards, for example Rf values for the the following carbohydrates are: Fructose = 0.68 Glucose = 0.64 Sucrose = 0.62 Maltose = 0.50 Lactose = 0.46

C

Solvent front

1. Calculate the Rf value for spot X in the example given above left (show your working):

2. Why is the Rf value of a substance always less than 1?

3. Predict what would happen if a sample was immersed in the chromatography solvent, instead of suspended above it:

(b) Based on the Rf values calculated in (a), identify the two amino acids:

Amino acid A:

Amino acid B:

5. (a) Which two amino acids could be difficult to separate by chromatography?

(b) Explain your reasoning:

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4. (a) Calculate the Rf values for the two unknown amino acid samples (above right):

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10 Protein Structure is Hierarchical

Key Idea: The sequence and type of amino acids in a protein determine its three-dimensional shape and function. Proteins are large, complex macromolecules, built up from a linear sequence of repeating units called amino acids. Proteins account for more than 50% of the dry weight of most cells and are important in virtually every cellular process. The

various properties of the amino acids, which are conferred by the different R groups, determine how the polypeptide chain folds up. This three dimensional tertiary structure gives a protein its specific chemical properties. If a protein loses this precise structure (through denaturation), it is usually unable to carry out its biological function.

Primary (1°) structure (amino acid sequence) Glu

Tyr

Ser

Iso

Met

Ala

Ala

Ser

Glu

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Phe

1. Describe the main features in the formation of each part of a protein's structure: (a) Primary structure:

Peptide bond

Amino acid

Hundreds of amino acids are linked by peptide bonds to form polypeptide chains. The attractive and repulsive charges on the amino acids determine the higher levels of organisation in the protein and its biological function.

(b) Secondary structure:

(c) Tertiary structure:

(d) Quaternary structure:

Secondary (2°) structure (a-helix or b pleated sheet)

Secondary (2°) structure is maintained by hydrogen bonds between neighbouring CO and NH groups. The hydrogen bonds are individually weak but collectively strong.

a-helix

Hydrogen bonds

Amino acid chain

b-pleated sheet

Polypeptide chains fold into a secondary (2°) structure based on H bonding. The coiled a-helix and b-pleated sheet are common 2° structures. Most globular proteins contain regions of both 2° configurations.

Tertiary (3°) structure (folding of the 2° structure)

Tertiary (3°) structure is maintained by more distant interactions such as disulfide bridges between cysteine amino acids, ionic bonds, and hydrophobic interactions.

a-helix

Aspartic acid

2. How are proteins built up into a functional structure?

Ionic bond Lysine

Disulfide bond

A protein's 3° structure is the three-dimensional shape formed when the 2° structure folds up and more distant parts of the polypeptide chains interact.

Quaternary (4°) structure

Alpha chain

Beta chain

A protein's 4° structure describes the arrangement and position of each of the subunits in a multiunit protein. The shape is maintained by the same sorts of interactions as those involved in 3° structure. ©2020 BIOZONE International ISBN: 978-1-98-856647-4 Photocopying Prohibited

3. Strong chemicals and extremes of temperature or pH can disrupt the bonds in proteins. What would this do to the protein's function and why?

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Prosthetic (haem) group

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Some complex proteins are only functional when as a group of polypeptide chains. Haemoglobin has a 4° structure made up of two alpha and two beta polypeptide chains, each enclosing a complex iron-containing prosthetic (or haem) group.

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11 Protein Shape is Related to Function

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Key Idea: The three dimensional shape of a protein reflects its role. When a protein is denatured, it loses its functionality. As we have seen, a protein may consist of one polypeptide chain, or several polypeptide chains linked together. Hydrogen bonds between amino acids cause the polypeptide chain to form its secondary structure, either an a-helix or a

b-pleated sheet. The interaction between R groups causes a polypeptide to fold into its tertiary structure, a three dimensional shape held by ionic bonds and disulfide bridges (bonds formed between sulfur containing amino acids). If bonds are broken (through denaturation), the protein loses its tertiary structure, and its functionality. H2N

The shape of a protein reflects its biological role

Phe Val

Leu

Ala

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Asn

Glu

Tyr

b sheets

Active site

Cys

Gly

Arg

a helix

Ser

H2N

S

Asn

COOH

Phe

Leu

Ile

Tyr

Val

Asn

Phe

Glu

Gln

a chain

Leu

Thr

Cys

S S

Cys Thr

S

Glu

Tyr

Cys

Gly

Gly

Cys

His

His

S

Gly

b chain

Leu

Val

Glu

Gln

Val

Leu

Gln

Pro

Tyr

Ser

S

Ile

Cys

Leu

Ser

Lys

Amylase

Thr

COOH

Channel proteins

Enzymes

Sub-unit proteins

Proteins that fold to form channels in the plasma membrane present non-polar R groups to the membrane and polar R groups to the inside of the channel. Hydrophilic molecules and ions are then able to pass through these channels into the interior of the cell. Ion channels are found in nearly all cells and many organelles.

Enzymes are globular proteins that catalyse specific reactions. Enzymes that are folded to present polar R groups at the active site will be specific for polar substances. Non-polar active sites will be specific for non-polar substances. Alteration of the active site by extremes of temperature or pH cause a loss of function.

Many proteins, e.g. insulin and haemoglobin, consist of two or more subunits in a complex quaternary structure, often in association with a metal ion. Active insulin is formed by two polypeptide chains stabilised by disulfide bridges between neighbouring cysteines. Insulin stimulates glucose uptake by cells.

Protein denaturation

When the chemical bonds holding a protein together become broken the protein can no longer hold its three dimensional shape. This process is called denaturation, and the protein usually loses its ability to carry out its biological function.

There are many causes of denaturation including exposure to heat or pH outside of the protein's optimum range. The main protein in egg white is albumin. It has a clear, thick fluid appearance in a raw egg (right). Heat (cooking) denatures the albumin protein and it becomes insoluble, clumping together to form a thick white substance (far right).

Cooked (denatured) egg white

Raw (native) egg white

1. Using the example of insulin, explain how interactions between R groups stabilise the protein's functional structure:

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3. Why does denaturation often result in the loss of protein functionality?

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2. Why do channel proteins often fold with non-polar R groups to the channel's exterior and polar R groups to its interior?

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12 Comparing Globular and Fibrous Proteins

Key Idea: The very different structure and properties of globular and fibrous proteins reflect their contrasting roles. Proteins can be classified according to structure or function. Globular proteins are spherical and soluble in water (e.g.

enzymes). Fibrous proteins have an elongated structure and are not water soluble. They provide stiffness and rigidity to the more fluid components of cells and tissues and have important structural and contractile roles. Fibrous proteins

Properties • Easily water soluble • Tertiary structure critical to function • Polypeptide chains folded into a spherical shape

Properties • Water insoluble • Very tough physically; may be supple or stretchy • Parallel polypeptide chains in long fibres or sheets

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Globular proteins

Function

• Catalytic, e.g. enzymes • Regulatory, e.g. hormones (insulin) • Transport, e.g. haemoglobin • Protective, e.g. immunoglobulins (antibodies)

Function

• Structural role in cells and organisms e.g. collagen in connective tissues, skin, and blood vessel walls. • Contractile e.g. myosin, actin The collagen molecule consists of three polypeptides wound together to form a helical ‘rope’. Every third amino acid in each polypeptide is a glycine (Gly) where hydrogen bonding holds the three strands together. Collagen molecules self assemble into fibrils held together by covalent cross linkages (below). Bundles of fibrils form fibres.

Hydrogen bond

Glycine

RuBisCO

Insulin

Insulin is a peptide hormone involved in the regulation of blood glucose. Insulin is composed of two peptide chains linked together by two disulfide bonds.

RuBisCo is a large multiunit enzyme. It catalyses the first step of carbon fixation in photosynthesis. It consists of 8 large and 8 small subunits and is the most abundant protein on Earth.

Haemoglobin is an oxygentransporting protein found in vertebrate red blood cells. One haemoglobin molecule consists of four polypeptide subunits. Each subunit contains an iron-containing haem group, which binds oxygen.

Many collagen molecules form fibrils and the fibrils group together to form larger fibres.

Collagen fibres

Haemoglobin

Covalent cross links between the collagen molecules

Rhinoceros horn is keratin

Collagen is the main component of connective tissue, and is mostly found in fibrous tissues (e.g. tendons, ligaments, and skin). Keratin is found in hair, nails, horn, hooves, wool, feathers, and the outer layers of the skin. The polypeptide chains of keratin are arranged in parallel sheets held together by hydrogen bonding.

(a) Structural tissues of the body:

(b) Catalysing metabolic reactions in cells:

2. How does the shape of a fibrous protein relate to its functional role?

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3. How does the shape of a catalytic protein (enzyme) relate to its functional role?

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1. How are proteins involved in the following roles? Give examples to help illustrate your answer:

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13 Protein Functional Diversity out most of a cell's work. A cell produces many different types proteins, each with a specific task. Proteins have roles in structure, function, and regulation of the body’s cells, tissues, and organs. Without a full complement of functional proteins, a cell can not carry out its specialised role. All of the proteins encoded by an organism's DNA is called its proteome. The proteome is larger than the genome because cells may produce many different proteins from one set of instructions.

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Key Idea: The order of nucleotides in DNA determines a protein's structure. Proteins carry out the essential functions of life and have structural, catalytic, and regulatory roles. In eukaryotic cells, most of a cell's genetic information (DNA) is found in a large membrane-bound organelle called the nucleus. DNA provides the instructions that code for the formation of proteins and the nucleus directs all cellular activities by controlling the synthesis of proteins, which carry

The nucleus is the control centre of a cell

The genome of an organism is its complete set of DNA, including all of its genes. Genes are different sections of DNA that encode specific proteins. A cell controls the number of different proteins it produces by expressing genes only as their proteins are required. This process of gene expression involves transcribing (rewriting) the gene and then translating it into a protein using the cell's molecular machinery.

The DNA within the nucleus provides instructions to a cell on how to carry out its functions to sustain essential life processes. This includes the production of proteins.

The nuclear envelope is formed by a doublelayered membrane. It keeps the DNA within the nucleus.

Nucleus

In eukaryotes, production of the protein is completed outside of the nucleus. Synthesis continues on ribosomes, which may be free in the cytoplasm or associated with the rough endoplasmic reticulum (rER).

An animal cell

1. (a) From the following word list match the correct function to the protein description below; internal defence, contractile, catalytic, regulation, structural, transport.

(b) From the following word list match the correct example to the protein description below: IgA, actin, amylase, insulin, haemoglobin, collagen, myosin, elastin, transport proteins in membranes.

Protein description

Function

Example(s)

Some proteins can function as enzymes, thereby controlling metabolism.

Some proteins can act as carrier molecules, to transport molecules from one place to another. Some proteins form antibodies that combat disease causing organisms. Some proteins form contractile elements in cells and bring about movement. WEB

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Proteins can make up structural components of tissues and organs.

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Proteins can function as chemical messenger molecules.

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Antibody

Antibodies (also called immunoglobulins) are "Y" shaped proteins that protect the body by identifying and killing disease-causing organisms such as bacteria and viruses.

EXAMPLE: IgA is an antibody found in the gut and airways. It destroys diseasecausing organisms growing in these areas and stops them causing an infection.

Movement

EXAMPLE: Actin and myosin are two proteins involved in the contraction of skeletal muscles.

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Cell

Internal defence

Lusb cc3.0

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Virus

Contraction

Enzyme catalyses breakdown of substrate

Contractile proteins are involved in movement of muscles and form the cytoskeleton of cells.

Catalytic

EXAMPLE: Digestion requires enzymes to break down food into smaller components. Amylase will break down the starch in this apple into maltose and glucose.

Thousands of different chemical reactions take place in an organism. Each chemical reaction is catalysed by enzymes. The ending "ase" identifies a molecule as an enzyme.

Regulation

Signalling protein

EXAMPLE: The hormone insulin is released after eating in response to high blood glucose and stimulates glucose uptake by cells. When blood glucose falls, the hormone glucagon stimulates processes that release glucose into the blood.

Regulatory proteins such as hormones act as signal molecules to control the timing and occurrence of biological processes and coordinate responses in cells, tissues, and organs.

Receptor protein

Structural

O2 Red blood cell Hb binds O2

Proteins can carry substances around the body or across membranes. For example, haemoglobin (Hb) transports oxygen (left) and proteins in cell membranes help molecules move into and out of cells.

EXAMPLE: Haemoglobin is a four unit protein found in red blood cells (left). It binds oxygen and carries it through the blood, delivering it to cells.

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Transport

O2 released

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Collagen fibre

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EXAMPLE: Collagen (left) is found in skin and connective tissues, including bones, tendons, and ligaments. Elastin is a highly elastic protein. Elastin helps skin to return to its original position when it is pinched.

Structural proteins provide physical support or protection. They are strong, fibrous (thread like) and stringy.


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14 How Are Proteins Modified?

Key Idea: The modification of proteins allows the cell to specify their use and final destination. Proteins may be modified after they have been produced by post translational modification. Two important modifications involve adding carbohydrates or lipids to the protein. Glycoproteins are formed by adding carbohydrates to proteins as they pass through the rough endoplasmic reticulum (rER) and Golgi. The carbohydrates may help

position and orientate the glycoprotein in the membrane, guide a protein to its final destination, or help in cell-to-cell recognition and cell signalling. Other proteins may have fatty acids added to them in the rER to form lipoproteins. These modified proteins transport lipids in the plasma between various organs in the body (e.g. gut, liver, and adipose tissue). Other common post-translational modifications include degradation, cleavage, and phosphorylation (below).

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Cleaving: Polypeptide chains may be cleaved to give smaller chains, which then fold or join to give the functional protein. An example is human insulin which is transcribed as one long polypeptide chain before being cleaved in two places to form two shorter chains that form the functional protein.

Glycosylation (adding carbohydrate groups): This is used to add an ID tag to the protein that will allow the cell to recognise its use and where it is to be transported (2a). The resulting glycoprotein may be used in the cell membrane or secreted. The carbohydrate tag may help position the glycoprotein within the membrane (2b).

P

Phosphorylation (the addition of phosphate groups) takes place in the Golgi. It may contribute to the protein's three dimensional structure or help with cell signalling.

P

P

P

Lipid attachment: Proteins may have lipids attached to them which anchor the protein to the plasma membrane.

Degradation: Some polypeptide chains may be tagged for degradation when they are no longer useful and their amino acids reused in the formation of other proteins.

(b) Why are these changes necessary?

2. Why might the orientation of a protein in the plasma membrane be important?

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1. (a) Describe some of the modifications that polypeptide chains undergo before becoming functional proteins:

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15 How Do We Know? DNA Carries the Code function suggested they could account for the many traits we see in organisms. Two early experiments, one by Griffith and another by Avery, MacLeod, and McCarty, provided important information about how traits could be passed on and what cellular material was responsible. The experiments involved strains of the bacterium Streptococcus pneumoniae. The S strain is pathogenic (causes disease). The R strain is harmless. Later experiments by Hershey and Chase helped confirm DNA as the genetic material.

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Key Idea: The progressive studies of several scientists helped to establish DNA as the heritable material responsible for the characteristics we see in organisms. Many years before Watson and Crick discovered the structure of DNA, biologists had deduced, through experimentation, that DNA carried the information that was responsible for the heritable traits we see in organisms. Prior to the 1940s, it was thought that proteins carried the code. Little was known about nucleic acids and the variety of protein structure and

Griffith (1928)

``Griffith found that when he mixed heat-killed pathogenic bacteria with living harmless cells, some of the living cells became pathogenic. Moreover, the newly acquired trait of pathogenicity was inherited by all descendants of the transformed bacteria. He concluded that the living R cells had been transformed into pathogenic cells by a heritable substance from the dead S cells.

Living R cells (no capsule, harmless)

Mouse lives

Living S cells (capsule, pathogenic)

Mouse dies

Heat-killed S cells

Mouse lives

Heat-killed S cells and living R cells

Mouse dies Blood sample contained living S cells that could reproduce to yield more S cells.

Avery-MacLeod-McCarty (1944)

``What was the unknown transformation factor in Griffith's experiment? Avery designed an experiment to determine if it was RNA, DNA, or protein. He broke open the heat-killed pathogenic cells and treated samples with agents that inactivated either protein, DNA, or RNA. He then tested the samples for their ability to transform harmless bacteria.

Treat sample with enzyme that destroys RNA, protein, or DNA.

Add the treated samples to cultures of R bacterial cells

RNase

S and R strains

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Protease S and R strains

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DNase

R strain only

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Heat-killed S cells, remove lipids and carbohydrates, homogenise and filter

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Hershey and Chase (1952)

``The work of Hershey and Chase in 1952 followed the work of Avery and his colleagues and was instrumental in the acceptance of DNA as the hereditary material. Hershey and Chase worked on viruses that infect bacteria called phages. Phages are composed only of DNA and protein. When they infect, they inject their DNA into the bacteria, leaving their protein coat outside.

``Hershey and Chase had two batches of phage. Batch 1 phage were grown with radioactively labelled sulfur, which was incorporated into the phage protein coat. Batch 2 phage were grown with radioactively labelled phosphorus, which was incorporated into the phage DNA. The phage were mixed with bacteria, which they then infected. When the bacterial cells were separated from the liquid, Hershey and Chase studied the liquid and the bacteria cells to determine where the radioactivity was.

``Hershey and Chase were able to show that DNA is the only material transferred directly from the phage to the bacteria when the

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bacteria are infected by the viruses.

Protein

The radioactivity is in the liquid supernatant

Radioactive protein capsule (35S)

Power

Speed

BBT Inc.

Radioactive DNA (32P)

DNA

The radioactivity is in the pellet

Power

Speed

BBT Inc.

Blender separates phage outside the bacteria from the cells and their contents.

Cells and phage separated using centrifugation. Cells and their DNA form the pellet.

1. Griffith did not predict transformation in his experiment. What results was he expecting? Explain:

2. (a) What did Avery's experimental results show?

(b) How did Avery's experiment build on Griffith's findings?

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3. (a) How did the Hershey-Chase experiment provide evidence that nucleic acids, not protein are the hereditary material?

(b) What assumptions were made about the role of DNA in viruses and bacteria and the role of DNA in eukaryotes?

(c) How would the results of the experiment have differed if proteins carried the genetic information?

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NH

OH OH

Nucleotide Structure

16 Nucleotides

P O CH 2 H N

H

N

H

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O

N

O

N

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N

A

NH2

H three a base, a sugar, and a phosphate group. Key Idea: nucleic Nacids. A nucleotide is N H N H Nucleotides make H components: NH H H H 2 Nucleotides may contain one of five bases. The combination made up of a base, NH a sugar, and aN phosphate OH group. Phosphate: Links H of bases in the nucleotides making up DNA or RNA stores Nucleotides are Nthe building blocks of the nucleic acids DNA neighbouring sugars. H OOH Base: One of four types N the information controlling the cell's activity.possible. The bases in and RNA, which are involved in the transmission of inherited OH P O CH Guanine Adenine 2 The base Phosphate O Sugar DNA are the same asNRNA except that thymine (T) in information. Nucleotide derivatives, Base such as ATP and GTP, carries theDNA codedisgenetic are involved in energy transfers in cells. message in a nucleic O A nucleotide has replaced withNuracil (U) in RNA. N

N up

H

Pyrimidines

H

H

H

H

Phosphate

CH2OH

Sugar: One of two types: ribose

Phosphate groups are represented

OH

RNA and deoxyribose in DNA. OH by circles. H Along withinthe pentose

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NH2

CH3

H N

N

H

H

Phosphate H N

H

N

O

O

H

N

O

H

H

sugar they form the "backbone" of Sugar Base the DNA or RNA molecule.

Purines

N

H

N

N

H

O

N

H OH

H

H

H

Deoxyribose

A

Phosphate: Links neighbouring sugars.

Base: One of five bases possible. The base carries the coded genetic message in a nucleic acid.

NH2

N

H

NH2

OH

H

Thymine Cytosine Uracil Pyrimidines are single ringed bases. DNA contains the pyrimidines cytosine (C) and thymine (T). RNA contains the pyrimidines cytosine and uracil (U).

H

P

N

O

H

H N

HO

O

O

N

N

H

O

Guanine

Sugar: One of two types: ribose in RNA and deoxyribose in DNA.

Adenine

Purines are double ringed bases. Both DNA and RNA contain the purines adenine (A) and guanine (G).

Nucleotide derivatives

O

H N

CH3

A

H

NH2

O

H 3 phosphate groups OH

N

H

Adenine N

O

H

Thymine ATP

HHON

N

O

P

O

OH H

H

N

O

H

Ribose Cytosine

CH2OH

H

Sugars

O

H

H

OH

H

OH

CH2OH

H

H

H

H

OH

H

Deoxyribose

Uracil

OH

O

H

OH

Ribose

Nucleotides contain one of two different sorts of sugars. Deoxyribose sugar is only found in DNA. Ribose sugar is found in RNA.

2 phosphates

A

Formation of a nucleotide

Adenine

ADP+ Pi

Condensation

(water removed)

Ribose

ATP is a nucleotide derivative used to provide chemical energy for metabolism. It consists of an adenine linked to a ribose sugar and 3 phosphate groups. Energy is made available when a phosphate group is transferred to a target molecule. Other nucleoside triphosphates (NTPs) have similar roles.

F

Nucleotide formation

Inorganic phosphate

A

A

H2 O

H2 O

In formation of a nucleotide, a phosphoric acid and a base are chemically bonded to a sugar molecule by condensation reactions in which water is given off. The reverse reaction is hydrolysis.

(a) In DNA: (b) In RNA:

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1. List the nucleotide bases present:

3. How can simple nucleotide units combine to store genetic information?

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2. Name the sugar present: (a) In DNA: (b) In RNA:

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17 Nucleic Acids

Key Idea: Nucleic acids are macromolecules made up of long chains of nucleotides, which store and transmit genetic information. DNA and RNA are nucleic acids. DNA and RNA are nucleic acids involved in the transmission of inherited information. Nucleic acids have the capacity to store the information that controls cellular activity. The

central nucleic acid is called deoxyribonucleic acid (DNA). Ribonucleic acids (RNA) are involved in the ‘reading’ of the DNA information. All nucleic acids are made up of nucleotides linked together to form chains or strands. The strands vary in the sequence of the bases found on each nucleotide. It is this sequence which provides the ‘genetic instructions’ for the cell.

Nucleotides are joined by condensation polymerisation

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A condensation reaction joins two molecules Formation together with of a the loss of a water molecule. dinucleotide In the formation of nucleic acids, nucleotides are joined together into polymers through a condensation reaction between the phosphate of one nucleotide and the sugar of another. Water is released. Because of the way they are formed, nucleic acids are called condensation polymers.

A

H2O

T

5

4

3

C

DNA molecule

In RNA, uracil replaces thymine in the code.

C

T

A

C

G

A

A

The carbon atoms on the pentose sugar are labelled one to five. During DNA replication (when new DNA is made) new nucleotides are added to the 3' end (the third carbon) of the existing nucleotide chain. It is therefore said DNA replication works in the 5' to 3' direction.

DNA molecule

C

G

G

Deoxyribose sugar

Hydrogen bonds hold the two strands together.

T

Only certain bases can pair.

Ribose sugar

Ribonucleic acid (RNA) comprises a single strand of nucleotides linked together. Although it is single stranded, it is often found folded back on itself, with complementary bases joined by hydrogen bonds.

2

New nucleotides added to this end.

RNAmolecule molecule RNA

U

1

Space filling model

Symbolic representation

Deoxyribonucleic acid (DNA) comprises a double strand of nucleotides linked together. It is shown unwound in the symbolic representation (above left). The DNA molecule takes on a double helix shape as shown in the space filling model above right.

Double-stranded DNA

5'

3'

The double-helix structure of DNA is like a ladder twisted into a corkscrew shape around its longitudinal axis. It is ‘unwound’ here to show the relationships between the bases.  The DNA backbone is made up of alternating phosphate and sugar molecules, giving the DNA molecule an asymmetrical structure.

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3'

C

G

T

5'

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A

 The ends of a DNA strand are labelled the 5' (five prime) and 3' (three prime) ends. The 5' end has a terminal phosphate group (off carbon 5), the 3' end has a terminal hydroxyl group (off carbon 3).

A

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G

 The asymmetrical structure gives a DNA strand direction. Each strand runs in the opposite direction to the other.

 The way the pairs of bases come together to form hydrogen bonds is determined by the number of bonds they can form and the configuration of the bases.

C

T

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RNAs contain selfcomplementary sequences that allow parts of the RNA to pair with itself to form short helices joined by H bonds.

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RNAs are involved in decoding the genetic information in DNA, as messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA). RNA is also involved in modifying mRNA after transcription and in regulating translation.

Messenger RNA (above) is transcribed (written) from DNA. It carries a copy of the genetic instructions from the DNA to ribosomes in the cytoplasm, where it is translated into a polypeptide chain.

Transfer RNA (above) carries amino acids to the growing polypeptide chain. One end of the tRNA carries the genetic code in a threenucleotide sequence called the anticodon. The amino acid links to the 3' end of the tRNA.

Ribosomal RNA (above) forms ribosomes from two separate ribosomal components (the large and small subunits) and assembles amino acids into a polypeptide chain.

1. Label the following parts on the diagram of the double-stranded DNA molecule at the bottom of page 24: (a) Deoxyribose (b) Phosphate (c) Hydrogen bonds (d) Purine bases (e) Pyrimidine bases 2. (a) Use the diagrams opposite to explain the base-pairing rule that applies in double-stranded DNA:

(b) How is the base-pairing rule for RNA different?

(c) What is the purpose of the hydrogen bonds in double-stranded DNA?

3. Briefly describe the roles of RNA:

4. (a) If you wanted to use a radioactive or fluorescent tag to label only the RNA in a cell and not the DNA, what molecule(s) would you label?

(b) If you wanted to use a radioactive or fluorescent tag to label only the DNA in a cell and not the RNA, what molecule(s) would you label?

5. (a) Why do the DNA strands have an asymmetrical structure?

(b) What are the differences between the 5' and 3' ends of a DNA strand?

Sugar present Bases present Number of strands Relative length

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6. Complete the following table summarising the differences between DNA and RNA molecules:


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18 Creating a DNA Model

Key Idea: Nucleotides pair together in a specific way called the base pairing rule. In DNA, adenine always pairs with thymine, and cytosine always pairs with guanine. DNA molecules are double stranded. Each strand is made up

of nucleotides. The chemical properties of each nucleotide mean it can only bind with one other type of nucleotide. This is called the base pairing rule and is explained in the table below. This exercise will help you to learn this rule.

DNA base pairing rule

Chargaff's rules

Adenine

always pairs with

Thymine

Thymine

always pairs with

Adenine T

A

Cytosine

always pairs with

Guanine C

G

Guanine

always pairs with

Cytosine G

C

A

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Before Watson and Crick described the structure of DNA, an Austrian chemist called Chargaff analysed the base composition of DNA from a number of organisms. He found that the base composition varies between species but that within a species the percentage of A and T bases are equal and the percentage of G and C bases are equal. Validation of Chargaff's rules was the basis of Watson and Crick's base pairs in the DNA double helix model.

T

1. Cut around the nucleotides on page 27 and separate each of the 24 nucleotides by cutting along the columns and rows (see arrows indicating two such cutting points). Although drawn as geometric shapes, these symbols represent chemical structures. 2. Place one of each of the four kinds of nucleotide on their correct spaces below:

Place a cut-out symbol for thymine here

Thymine

Place a cut-out symbol for adenine here

Place a cut-out symbol for cytosine here

Cytosine

Place a cut-out symbol for guanine here

Adenine

Guanine

3. Identify and label each of the following features on the adenine nucleotide immediately above: phosphate, sugar, base, hydrogen bonds

4. Create one strand of the DNA molecule by placing the 9 correct 'cut out' nucleotides in the labelled spaces on the following page (DNA molecule). Make sure these are the right way up (with the P on the left) and are aligned with the left hand edge of each box. Begin with thymine and end with guanine. 5. Create the complementary strand of DNA by using the base pairing rule above. Note that the nucleotides have to be arranged upside down.

Factor 1:

Factor 2:

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6. Under normal circumstances, it is not possible for adenine to pair up with guanine or cytosine, nor for any other mismatches to occur. Describe the two factors that prevent a mismatch from occurring:

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7. Once you have checked that the arrangement is correct, you may glue, paste or tape these nucleotides in place.

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Cut

Cut

Cut

P

P

P

P

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P

P

S

S

S

S

S

S

Adenine

Guanine

Adenine

Guanine

Adenine

Guanine

P

P

P

P

P

P

S

S

S

S

S

S

Thymine

Cytosine

Thymine

Cytosine

Thymine

Cytosine

P

P

P

P

P

P

S

S

S

S

S

S

Adenine

Guanine

Adenine

Guanine

Adenine

Guanine

P

P

P

P

P

P

S

S

S

S

S

S

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Cytosine

Thymine

Cytosine

Thymine

Cytosine

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27

Nucleotides

Tear out this page and separate each of the 24 nucleotides by cutting along the columns and rows (see arrows indicating the cutting points).

Cut

Cut

Cut

Cut


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P

Adenine

S

Thymine S Thymine

Put the matching complementary nucleotides opposite the template strand

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Put the named nucleotides on the left hand side to create the template strand

DNA molecule

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Cytosine

Adenine

Adenine

Guanine

Thymine

Thymine

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Guanine

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Cytosine


19 How Does DNA Replicate?

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Key Idea: Semi conservative DNA replication produces two identical copies of DNA, each containing half original material and half new material. Before a cell can divide, it must double its DNA. It does this by a process called DNA replication. This process ensures that each resulting cell receives a complete set of genes from the

original cell. After the DNA has replicated, each chromosome is made up of two chromatids, joined at the centromere. The two chromatids will become separated during cell division to form two separate chromosomes. During DNA replication, nucleotides are added at the replication fork. Enzymes are responsible for all of the key events.

Step 1 Unwinding the DNA molecule

5'

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3'

A normal chromosome consists of an unreplicated DNA molecule. Before cell division, this long molecule of double stranded DNA must be replicated.

For this to happen, it is first untwisted and separated (unzipped) at high speed at its replication fork by an enzyme called helicase. Another enzyme relieves the strain that this generates by cutting, winding and rejoining the DNA strands.

Single-armed chromosome as found in a non-dividing cell.

Temporary break allows the strand to swivel

Free nucleotides are used to construct the new DNA strand.

Helicase at the replication fork

Step 2 Making new DNA strands

The formation of new DNA is carried out mostly by an enzyme complex called DNA polymerase.

DNA polymerase

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DNA polymerase catalyses the condensation reaction that joins adjacent nucleotides. The strand is synthesised in a 5' to 3' direction, with the polymerase moving 3' to 5' along the strand it is reading. Thus the nucleotides are assembled in a continuous fashion on one strand but in short fragments on the other strand. These fragments are later joined by an enzyme to form one continuous length.

Step 3 Rewinding the DNA molecule

Each of the two new double-helix DNA molecules has one strand of the original DNA (dark grey and white) and one strand that is newly synthesised (blue). The two DNA molecules rewind into their double-helix shape again.

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5'

The two new strands of DNA coil into a double helix

3'

5'

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Replicated chromosome ready for cell division.

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DNA replication is semi-conservative, with each new double helix containing one old (parent) strand and one newly synthesised (daughter) strand. The new chromosome has twice as much DNA as a non-replicated chromosome. The two chromatids will become separated in the cell division process to form two separate chromosomes.

Each of the newly formed DNA molecules create a chromatid.

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1. What is the purpose of DNA replication?

2. Summarise the three main steps involved in DNA replication: (a)

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(b)

(c)

3. For a cell with 22 chromosomes, state how many chromatids would exist following DNA replication: 4. State the percentage of DNA in each daughter cell that is new and the percentage that is original:

5. What does it mean when we say DNA replication is semi-conservative?

6. How are the new strands of DNA lengthened during replication:

7. What rule ensures that the two new DNA strands are identical to the original strand?

8. Why does one strand of DNA need to be copied in short fragments?

9. Match the statements in the table below to form complete sentences, then put the sentences in order to make a coherent paragraph about DNA replication and its role: DNA replication is the process by which the DNA molecule...

...by enzymes.

Replication is tightly controlled...

...to correct any mistakes.

After replication, the chromosome...

...and half new DNA.

DNA replication...

...during mitosis.

The chromatids separate...

...is copied to produce two identical DNA strands.

A chromatid contains half original ...

...is made up of two chromatids.

Write the complete paragraph here:

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...is required before mitosis or meiosis can occur.

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The enzymes also proofread the DNA during replication...


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20 Modeling DNA Replication

Key Idea: Meselson and Stahl devised an experiment that showed DNA replication is semi-conservative.

Models for DNA replication ffSeveral models were proposed for DNA

This practical activity models DNA replication. Use it to replicate Meselson and Stahl's DNA replication experiment.

Conservative model

replication (right).

ffThe conservative model proposed that when

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DNA is replicated one of the new DNA molecules receives two newly replicated DNA strands and the other receives the two original DNA strands.

Semi-conservative model

ffThe semi-conservative model proposed that

each DNA strand in the original molecule served as a template for the new DNA molecules. Therefore the two new DNA molecules each contained one strand of original DNA and one strand of new DNA.

Original DNA strand

New DNA strand

ffOn the next page you will model DNA replication.

Meselson and Stahl's experiment

Separating DNA by density

ffTwo scientists, called Meselson and Stahl, determined that DNA replication is semi-conservative.

ffThey grew bacteria in a solution containing heavy nitrogen (15N) until

their DNA contained only 15N. The bacteria were placed into a growth solution containing normal nitrogen (14N), which is lighter than 15N. After set generation times the DNA was extracted and centrifuged in a caesium chloride (CsCl) solution. The CsCl provides a density gradient to separate the DNA. Heavy DNA sinks to the bottom, light DNA rises to the top, and intermediate DNA (one light and one heavy strand) settles in between (right).

ffIf DNA replicated conservatively, after one generation there would be two kinds of DNA. Heavy DNA consisting of two stands of light DNA consisting of two 14N stands of DNA.

15N

DNA, and

Caesium chloride solution Light DNA

Intermediate DNA Heavy DNA

ffIf DNA replicated semi-conservatively, after one generation there would be one type of DNA, an intermediate density DNA consisting of one strand of heavy 15N and one strand of light 14N DNA. This is the result that Meselson and Stahl obtained.

Semi-conservative model

1. Why did Meselson and Stahl's experiment support the semiconservative replication model?

2. (a) Complete the diagram (right) for the next DNA replication.

(b) Predict what would happen to the proportion of intermediate DNA under the semi-conservative method in future generations:

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Conservative model

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Instructions

1. Cut out the DNA shapes provided on this page. 2. Intertwine the first pair (labelled 0) of heavy 15N (black) DNA. This forms Generation 0 (parental DNA). 3. Use the descriptions of the two possible models for DNA replication on the previous page to model semiconservative and conservative DNA replication.

2 2

2

2

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1

2

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2

0

1

0

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4. For each replication method, record on a separate sheet of paper the percentage of heavy 15N-15N (black-black), intermediate 15N-14N (black-grey), and light 14N-14N (grey-grey).


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21 The Evidence for DNA Structure

Key Idea: Many scientists contributed to the discovery of DNA's structure. Once the structure of DNA was known, it immediately suggested a mechanism for its replication. DNA is easily extracted and isolated from cells. This was first done in 1869, but it took the work of many scientists working

1951 Rosalind Franklin studied the structure of DNA using x-ray crystallography at King's College. Maurice Wilkins, also at King's College, was also using x-ray crystallography to study DNA. Wilkins and Franklin did not get on well.

30 January 1953 Wilkins shows Watson and Crick the "photo 51" without Franklin's approval or knowledge. It provided the structural information they needed to finalise their model, completed on 7 March 1953.

16 April 1958 Franklin dies at age 37 of ovarian cancer. She was never nominated for a Nobel Prize.

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Late 1940s Linus Pauling determined by x-ray crystallography that proteins have a helical structure.

in different areas many years to determine DNA's structure. In particular, four scientists, Watson, Crick, Franklin, and Wilkins are recognised as having made significant contributions in determining the structure of DNA. Once the structure of DNA was known scientists could determine how it was replicated.

1949 Chargaff’s rules are announced: DNA contains equal proportions of bases A and T and G and C.

1951 James Watson and Francis Crick build their first DNA model - a three stranded helix, with bases to the outside and phosphate groups to the inside. Franklin points out that their model is incorrect and is not consistent with the data.

May 1952 Franklin produces "photo 51", showing DNA is a helix. She was working on a less hydrated form of DNA and did not return to the photo again until 1953.

1962 Watson, Crick, and Wilkins win the Nobel Prize in Physiology or Medicine. Franklin was not acknowledged.

Discovering the structure of DNA ... a story of collaboration and friction

Although Watson and Crick are often credited with discovering DNA's structure, the contributions of many scientists were important and personal conflicts and internal politics probably prevented the structure being determined earlier. Professional friction between Franklin and Wilkins meant that they worked independently of each other. Watson and Crick analysed Franklin's results without her knowledge or consent and Watson himself recalls that he tended to dismiss her. Only later did he acknowledge her considerable contribution. Franklin did not receive the Nobel prize, which cannot be awarded posthumously.

A

A

Photo 51

Wilkins and Franklin were both crystallographers. Their X-ray diffraction patterns of DNA provided measurements of different parts of the molecule and the position of different groups of atoms. Franklin's X-ray image (photo 51) gave Watson and Crick the necessary information to produce their model of DNA. As soon as Franklin saw it, she readily accepted it as correct.

1. What made Watson and Crick realise that DNA was a double helix?

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Distinct parts of the famous "photo 51" X-ray diffraction image (recreated in the illustration right) indicate specific qualities of the DNA. The X pattern indicates a helix, but Watson and Crick realised that the apparent gaps in the X (labelled A) were due to the repeating pattern of a double helix. The diamond shapes (in blue) indicate the helix is continuous and of constant dimensions and that the sugar-phosphate backbone is on the outside of the helix. The distance between the dark horizontal bands allows the calculation of the length of one full turn of the helix.

2. Do you regard the discovery of DNA's structure as a collaboration between scientists working in related fields? Explain:

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22 What is Gene Expression?

Key Idea: Genes are sections of DNA that code for proteins. Genes are expressed when they are transcribed into messenger RNA (mRNA) and then translated into a protein. Gene expression is the process by which the information in a gene is used to synthesise a protein. It involves transcription of the DNA into mRNA and translation of the mRNA into

protein. Eukaryotic genes include non-protein coding regions called introns. These regions of intronic DNA must be edited out before the mRNA is translated by the ribosomes. Transcription of the genes and editing that primary transcript to form the mature mRNA occurs in the nucleus. Translation of the protein by the ribosomes occurs in the cytoplasm.

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A summary of eukaryotic gene expression

Nucleus

TRANSLATION

EDITING

TRANSCRIPTION

Ribosome

mRNA

Nuclear pore

Primary transcript

mRNA

Amino acids are linked together at the ribosome to form the protein encoded by mRNA.

Cytoplasm

The primary transcript is edited. The non-protein coding introns are removed and modifications are made to help the mRNA exit the nucleus.

DNA

In the nucleus, the gene is rewritten into a single stranded primary RNA transcript, using one strand of DNA as a template. RNA polymerase catalyses this process.

1. What is a gene?

2. (a) What does gene expression mean?

(b) What are the three stages in gene expression in eukaryotes and what happens in each stage?

(i) (ii)

(iii)

(a) What is the consequence of active transcription in a polytene chromosome?

(b) Why might this be useful in a larval insect?

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3. The photograph right shows an SEM of a giant polytene chromosome. These chromosomes are common in the larval stages of flies, which must grow rapidly before changing to the adult form. They form as a result of repeated cycles of DNA replication without cell division. This creates many copies of genes. Within these chromosomes, visible 'puffs' indicate regions where there is active transcription of the genes.

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23 The Genetic Code

Key Idea: The genetic code is the set of rules by which the genetic information in DNA or mRNA is translated into proteins. The genetic information for the assembly of amino acids is stored as three-base sequence. These three letter codes on mRNA are called codons. Each codon represents one of 20 amino acids used to make proteins. The code is effectively Codons that code for this amino acid

No.

Codons that code for this amino acid

Amino acid

No.

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Amino acid

universal, being the same in all living things (with a few minor exceptions). The genetic code is summarised in a mRNAamino acid table, which identifies the amino acid encoded by each mRNA codon. The code is degenerate, meaning there may be more than one codon for each amino acid. Most of this degeneracy is in the third nucleotide of a codon.

GCU, GCC, GCA, GCG

4

Leu

Leucine

Arginine

Lys

Lysine

Asn

Asparagine

Met

Methionine

Asp

Aspartic acid

Phe

Phenylalanine

Cys

Cysteine

Pro

Proline

Gln

Glutamine

Ser

Serine

Glu

Glutamic acid

Thr

Threonine

Gly

Glycine

Trp

Tryptophan

His

Histidine

Tyr

Tyrosine

Ile

Isoleucine

Val

Valine

Ala

Alanine

Arg

1. Use the mRNA-amino acid table (below) to list in the table above all the codons that code for each of the amino acids and the number of different codons that can code for each amino acid (the first amino acid has been done for you). 2. (a) How many amino acids could be coded for if a codon consisted of just two bases?

(b) Why is this number of bases inadequate to code for the 20 amino acids required to make proteins?

3. Describe the consequence of the degeneracy of the genetic code to the likely effect of a change to one base in a triplet:

C on the left row, A on the top column, G on the right row CAG is Gln (glutamine)

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U

C

A

G

U

UUU UUC UUA UUG

Phe Phe Leu Leu

UCU UCC UCA UCG

Ser Ser Ser Ser

UAU UAC UAA UAG

Tyr Tyr STOP STOP

UGU UGC UGA UGG

Cys Cys STOP Trp

U C A G

C

CUU CUC CUA CUG

Leu Leu Leu Leu

CCU CCC CCA CCG

Pro Pro Pro Pro

CAU CAC CAA CAG

His His Gln Gln

CGU CGC CGA CGG

Arg Arg Arg Arg

U C A G

A

AUU AUC AUA AUG

Ile Ile Ile Met

ACU ACC ACA ACG

Thr Thr Thr Thr

AAU AAC AAA AAG

Asn Asn Lys Lys

AGU AGC AGA AGG

Ser Ser Arg Arg

U C A G

G

GUU GUC GUA GUG

Val Val Val Val

GCU GCC GCA GCG

Ala Ala Ala Ala

GAU GAC GAA GAG

Asp Asp Glu Glu

GGU GGC GGA GGG

Gly Gly Gly Gly

U C A G

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Example: Determine CAG

Read third letter here

Second letter

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Third letter

How to read the table: The table on the right is used to 'decode' the genetic code as a sequence of amino acids in a polypeptide chain, from a given mRNA sequence. To work out which amino acid is coded for by a codon (triplet of bases) look for the first letter of the codon in the row label on the left hand side. Then look for the column that intersects the same row from above that matches the second base. Finally, locate the third base in the codon by looking along the row from the right hand end that matches your codon.

Read second letter here

First letter

mRNA-amino acid table

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24 Cracking the Genetic Code

Key Idea: Scientists used mathematics and experiments to unlock the genetic code. A series of three nucleotides, called a triplet, codes for a single amino acid. In the 1960s, two scientists, Marshall Nirenberg and Heinrich

Matthaei, developed an experiment to crack the genetic code. Their experiment, which is sometimes called the poly-U experiment because it used a synthetic mRNA containing only uracil, is shown below.

How was the genetic code cracked?

Once it was discovered that DNA carries the genetic code needed to produce proteins, the race was on to "crack the code" and find out how it worked.

Once the triplet code was discovered, the next step was to find out which amino acid each codon produced. Two scientists, Marshall Nirenberg and Heinrich Matthaei, developed an experiment (below) to crack the code.

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The genetic code

The first step was to find out how many nucleotide bases code for an amino acid. Scientists knew that there were four nucleotide bases in mRNA, and that there are 20 amino acids. Simple mathematics (below) showed that a one or two base code did not produce enough amino acids, but a triplet code produced more amino acids than existed. The triplet code was accepted once scientists confirmed that some amino acids have multiple codes.

Number of bases in the code

Working

Number of amino acids produced

Single (41)

4

4 amino acids

Double (42)

4x4

16 amino acids

Triple (43)

4x4x4

64 amino acids

A triplet (three nucleotide bases) codes for a single amino acid. The triplet code on mRNA is called a codon.

A cell free E. coli extract was produced for their experiment by rupturing the bacterial cells to release the cytoplasm. The extract had all the components needed to make proteins (except mRNA).

1 DNase was

added to destroy bacterial DNA so there was no template for mRNA to be made.

2 Radio labelled amino

acids and a synthetic mRNA strand containing only uracil (U) were added.

U U U U U U U U U U U U

Cell free E. coli extract.

3 Once the mRNA was

Phe Phe Phe Phe

added an amino acid was produced. The codon UUU produced the amino acid phenylalanine (Phe).

4 Over the next few years, similar

experiments were carried out using different combinations of nucleotides until all of the codes were known.

1. (a) How many types of nucleotide bases are there in mRNA?

(b) How many types of amino acids are there in proteins?

(c) Why did scientists reject a one or two base code when trying to work out the genetic code?

(b) What would it have been difficult to crack the code if no DNase was added?

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3. (a) Why was DNase added to the cell free E. coli extract?

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2. A triplet code could potentially produce 64 amino acids. Why are only 20 amino acids produced?

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25 Transcription in Eukaryotes

Key Idea: Transcription is the first step of gene expression. It involves the enzyme RNA polymerase rewriting the information into a primary RNA transcript. In eukaryotes, transcription takes place in the nucleus. Transcription is the first stage of gene expression. It takes place in the nucleus and is carried out by the enzyme RNA polymerase, which rewrites the DNA into a primary RNA transcript using a single template strand of DNA. The

protein-coding portion of a gene is bounded by an upstream start (promoter) region and a downstream terminator region. These regions control transcription by telling RNA polymerase where to start and stop transcription. In eukaryotes, non protein-coding sections called introns must first be removed and the remaining exons spliced together to form the mature mRNA before the gene can be translated into a protein. This editing process also occurs in the nucleus.

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Transcription is carried out by RNA polymerase (RNAP)

3'

RNA polymerase (RNAP) adds nucleotides to the 3' end so the strand is synthesised in a 5' to 3' direction.

mRNA nucleotides. Free nucleotides are used to construct the RNA strand

3'

Template (antisense) strand of DNA stores the information that is transcribed into mRNA

Direction of transcription

RNA polymerase binds at the upstream promoter region

Newly synthesised RNA strand is complementary to the template strand

RNA polymerase dissociates at the terminator region

3'

5'

The primary RNA transcript is edited to form the mature mRNA and then passes to the cytoplasm where the nucleotide sequence is translated into a polypeptide.

Translation will begin at the start codon AUG

5'

Several RNA polymerases may transcribe the same gene at any one time, allowing a high rate of mRNA synthesis.

1. (a) Name the enzyme responsible for transcribing the DNA: (b) What strand of DNA does this enzyme use?

(c) The code on this strand is the [ same as / complementary to ] the RNA being formed (circle correct answer).

(d) Which nucleotide base replaces thymine in mRNA?

(e) On the diagram, use a coloured pen to mark the beginning and end of the protein-coding region being transcribed.

2. (a) In which direction is the RNA strand synthesised?

(b) Explain why this is the case:

3. (a) Why is AUG called the start codon?

(b) What would the three letter code be on the DNA coding strand?

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5'

Coding (sense) strand of DNA

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40 Comparing gene expression in prokaryotes and eukaryotes

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In both prokaryotes and eukaryotes, genes are transcribed by RNA polymerase and translated by ribosomes. However, there are some important differences. In eukaryotes, primary RNA must be edited and processed before passing from the nucleus to the cytoplasm. In prokaryotes, there is no nucleus and ribosomes can begin translating a gene while it is still being transcribed.

Nucleus

DNA

Cytoplasm

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DNA

RNAP

Transcription

5'

mRNA

RNAP

Transcription and processing 3'

mRNA

Transport

5'

Translation by ribosomes

3'

Translation by ribosomes

5'

Growing polypeptide

Prokaryote

Eukaryote

4. For the following triplets on a DNA template strand, state the codon sequence for the mRNA that would be synthesised:

Triplets on the DNA:

T A C

Codons on the mRNA: Triplets on the DNA:

T A C

T A G

C C G

C G A

T T T

A A G

C C T

A T A

A A A

Codons on the mRNA: 5. What is the significance of the promoter and terminator regions on the DNA?

6. Why might a cell employ several RNA polymerases to produce multiple RNA transcripts of a gene at any one time?

7. Based on the diagram above, describe two differences between gene expression in prokaryotes and eukaryotes: (a)

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(b)

8. Based on the diagram above, how is gene expression in prokaryotes and eukaryotes similar?

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9. What benefit might there be to the way in which prokaryotes can begin translation while a gene is still being transcribed?

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26 mRNA Processing in Eukaryotes

Key Idea: Primary mRNA molecules are modified after transcription so that the mRNA can exit the nucleus. Post transcriptional modification also enables the cell to produce a wide variety of proteins from a smaller number of genes. Once a gene is transcribed, the primary transcript is modified to produce the mRNA strand that will be translated in the

cytoplasm. Modifications to the 5' and 3' ends of the transcript enable the mRNA to exit the nucleus and remain stable long enough to be translated. Other post transcriptional modifications remove non-protein coding intronic DNA and splice exons in different combinations to produce different protein end products.

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Primary RNA is modified by the addition of caps and tails ``After transcription, both ends of the primary RNA are modified by enzymes to create 'caps' and 'tails' (below). These modifications protect the RNA from degradation and help its transport through the nuclear pore. Primary RNA

5' modification (capping) A guanine nucleotide cap is added to the 5' end of the primary transcript to protect it from degradation during transport from the nucleus to the cytoplasm.

3' modification (poly-A tails) Multiple adenosine nucleotides are added to the primary transcript. These poly-A tails aid nuclear export, translation, and stability of the mRNA.

Post transcriptional modification ``Human DNA contains 25,000 genes, but produces up to one million different proteins. Each gene must therefore produce more than one protein. This is achieved through both post-transcriptional modification of the mRNA as well as post translational modifications, such as glycosylation and addition of phosphates.

``Primary RNA contains both protein coding exons and non-protein coding introns. Introns are usually removed after transcription and may be processed to create regulatory elements such as microRNAs. The exons are then spliced together ready to be translated. However, there are many alternative ways to splice the exons and these alternatives create variations in the translated proteins. The most common method of alternative splicing involves exon skipping, in which not all exons are spliced into the final mRNA create further variants. Exon 1 (below). Other splicing Exonoptions 2 Exon 3 Exon 4 Exon 5

Exon 1

Exon 2

Intron

Exon 3

Exon 4

Intron

Exon 5

Intron

Intron

Splicing

1

2

3

4

5

1

2

4

5

1

2

3

5

1

2

3

4

5

1

2

4

5

1

2

3

5

Three splicing alternatives creates three different proteins

1. What is the purpose of the caps and tail on mRNA?

(b) What is one possible fate for these introns?

3. How can so many proteins be produced from so few genes?

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2. (a) What happens to the intronic sequences in DNA after transcription?

4. If a human produces 1 million proteins, but human DNA codes for only 25,000 genes, on average how many proteins are produced per gene?

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27 Translation

Key Idea: Translation is the final stage of gene expression in which ribosomes read the mRNA and decode (translate) it to synthesise a protein. This occurs in the cytoplasm. In eukaryotes, translation occurs in the cytoplasm either at free ribosomes or ribosomes on the rough endoplasmic reticulum. Ribosomes translate the code carried in the mRNA

molecules, providing the catalytic environment for the linkage of amino acids delivered by transfer RNA (tRNA) molecules. Protein synthesis begins at the start codon and, as the ribosome wobbles along the mRNA strand, the polypeptide chain elongates. On reaching a stop codon, the ribosome subunits dissociate from the mRNA, releasing the protein.

tRNA structure

Ribosomes are made up of a complex of ribosomal RNA (rRNA) and ribosomal proteins. These small cellular structures direct the catalytic steps required for protein synthesis and have specific regions that accommodate transfer RNA (tRNA) molecules loaded with amino acids.

tRNA molecules are RNA molecules, about 80 nucleotides long, which transfer amino acids to the ribosome as directed by the codons in the mRNA. Each tRNA has a 3-base anticodon, which is complementary to a mRNA codon. There is a different tRNA molecule for each possible codon and, because of the degeneracy of the genetic code, there may be up to six different tRNAs carrying the same amino acid.

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Ribosome structure

Ribosomes exist as two separate sub-units (below) until they are attracted to a binding site on the mRNA molecule, when they come together around the mRNA strand.

Amino acid attachment site. Enzymes attach the tRNAs to their specific amino acids.

Large subunit

Large subunit

Anticodon is a 3-base sequence complementary to the codon on mRNA.

subunit Functional Small ribosome

1. Describe the structure of a ribosome:

2. What is the role of each of the following components in translation?

(a) Ribosome:

(b) tRNA:

(c) Amino acids:

(d) Start codon:

(e) Stop codon:

3. There are many different types of tRNA molecules, each with a different anticodon (HINT: see the mRNA table). (a) How many different tRNA types are there, each with a unique anticodon?

(b) Explain your answer:

(c) Determine the mRNA codons and the amino acid sequence for the following tRNA anticodons:

Codons on the mRNA:

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U A C

U A G

C C G

C G A

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tRNA anticodons:

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43 tRNA molecules deliver amino acids to ribosomes

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tRNA molecules match amino acids with the appropriate codon on mRNA. As defined by the genetic code, the anticodon specifies which amino acid the tRNA carries. The tRNA delivers its amino acid to the ribosome, where enzymes join the amino acids to form a polypeptide chain. During translation the ribosome "wobbles" along the mRNA molecule joining amino acids together. Enzymes and energy are involved in charging the tRNA molecules (attaching them to their amino acid) and elongating the peptide chain.

Unloaded Met-tRNA

Lys Charged Arg-tRNA enters the ribosome A (acceptor) site. The amino acid is added to the growing polypeptide chain.

Charging Lys-tRNA

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The ribosome P (peptidyl) site carries the growing polypeptide chain.

Met

Thr

Phe

Unloaded Thr-tRNA leaves the ribosome E (exit) site

Charged Val-tRNA

Val

Arg

Charged tRNAs enter at the A site except for the first amino acid methionine (Met), which enters at the P site to begin the process.

Start codon

5'

3'

mRNA

Ribosome (only large subunit shown)

The polypeptide chain grows as more amino acids are added The polypeptide chain continues to grow as more amino acids are added.

Protein synthesis stops when a stop codon is reached (UGA, UAA, or UAC). The ribosome falls off the mRNA and the polypeptide is released.

Protein synthesis begins when the ribosome reads the start codon (AUG).

START

STOP

5'

3'

Direction of protein synthesis

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4. Describe the events occurring during translation:

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5. Many ribosomes can work on one strand of mRNA at a time (a polyribosome system). What would this achieve?


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28 Gene Expression Summary 3'

Thr Phe

Cys

Asn

Lys

Met

Lys

Tyr

Arg Val Met Tyr

Thr Phe Arg Val

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5'

Lys

Tyr

Met

Arg

Thr

Thr

Ala

Phe

Gene #2

5'

Gene #1

3'

This diagram provides a visual overview of gene expression. It combines information from the previous activities. Each of the major steps in the process are numbered, whereas structures are identified with letters.

1. Briefly describe each of the numbered processes in the diagram above:

(a) Process 1:

(b) Process 2:

(c) Process 3:

(d) Process 4:

(e) Process 5:

(f) Process 6:

(g) Process 7:

(a) Structure A:

(f) Structure F:

(b) Structure B:

(g) Structure G:

(c) Structure C:

(h) Structure H:

(d) Structure D:

(i) Structure I:

(e) Structure E:

(j) Structure J:

(b)

WEB

REVISE

28

LINK

25

LINK

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3. Describe two factors that would determine whether or not a particular protein is produced in the cell: (a)

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(h) Process 8: (i) Process 9: 2. Identify each of the structures marked with a letter and write their names below in the spaces provided:

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29 KEY TERMS AND IDEAS: Did You Get It?

1. Test your vocabulary by matching each term to its correct definition, as identified by its preceding letter code.

base-pairing rule coding

strand

B A set of rules by which information encoded in DNA or mRNA is translated into proteins. C The rule governing the pairing of complementary bases in DNA. D Two-ringed organic bases, including adenine and guanine.

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DNA

A Single stranded nucleic acid that consists of nucleotides containing ribose sugar.

genetic code

hydrogen bonding nucleic acids nucleotides

E Form of intermolecular bonding between hydrogen and an electronegative atom such as oxygen. F The sequence of DNA that is read during the synthesis of mRNA.

G Universally found macromolecules composed of chains of nucleotides. These molecules carry genetic information within cells.

purines

H The DNA strand with the same base sequence as the RNA transcript produced (although with thymine replaced by uracil in mRNA).

pyrimidines

I

RNA

template strand

Macromolecule consisting of many millions of units containing a phosphate group, sugar and a base (A,T, C or G). Stores the genetic information of the cell.

J The structural units of nucleic acids, DNA and RNA.

K Single-ringed organic bases, including uracil, cytosine, and thymine.

2. (a) On the diagram shown right, highlight the structure that indicates a DNA helix.

(b) Circle the region that indicates there is a double helix.

(c) What do the blank diamond shaped areas in the diagram indicate?

3. For the following DNA sequence on the template strand, give the mRNA sequence and then Identify the amino acids that are encoded. For this question you may consult the mRNA-amino acid table earlier in the chapter.

DNA (template strand): G A A A C C C T T A C A T A T C G T G C T

mRNA:

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Amino acids: 4. Complete the following paragraph by deleting one of the words in the bracketed () pairs below:

In eukaryotes, gene expression begins with (transcription/translation) which occurs in the (cytoplasm/nucleus).

(Transcription/Translation) is the copying of the DNA code into (mRNA/tRNA). The (mRNA/tRNA) is then transported to

the (cytoplasm/nucleus) where (transcription/translation) occurs. Ribosomes attach to the (mRNA/tRNA) and help match

the codons on (mRNA/tRNA) with the anticodons on (mRNA/tRNA). The (mRNA/tRNA) transports the animo acids to the

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ribosome where they are added to the growing (polypeptide/carbohydrate) chain.

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Gene structure and regulation

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Unit 3 Outcome 1

Key terms

Structural and regulatory genes

exon

Key knowledge

gene expression gene induction

gene repression

Activity number

c

1

Distinguish between the function of structural and regulatory genes.

c

2

Where are regulatory genes located in relation to the genes they control? Is this different for prokaryotes and eukaryotes?

30

inducible genes

Eukaryotic gene structure

intron

lac operon

Key knowledge

operator operon

c

3

Describe the structure of eukaryotic genes with reference to the start and stop (terminator) signals for transcription, the promoter region, and exons and introns. Distinguish between the start and stop signals for transcription and the start and stop codons for translation.

c

4

Explain how transcription in eukaryotes involves the interaction of transcription factors with specific sequences so that RNA polymerase can transcribe (rewrite) the protein-coding gene into a primary RNA transcript. Recall that this primary transcript is edited to form the mature mRNA.

promoter

regulatory gene repressor

Activity number

structural gene TATA box

30 31

31

terminator sequence transcription factors

Matthias Zepper

Prokaryotic gene structure and regulation Key knowledge

Activity number

5

Describe the operon model of gene expression in prokaryotes. Explain how the structural genes in an operon are switched on or off by proteins expressed by regulatory genes.

32

c

6

Describe the role of the regulator gene in producing the repressor molecule and explain what the repressor molecule does.

32

c

7

Recognise that both gene induction and gene repression are involved in the regulation of gene expression in bacteria.

32

c

8

Explain gene induction in prokaryotes, as illustrated by the lac operon in E. coli in which lactose (an uncommon substrate) acts as an inducer for transcription of the structural genes for lactose metabolism.

32

c

9

Explain gene repression in prokaryotes in which the transcription of genes that are normally transcribed all the time are switched off. Know that some genes (e.g. ribosomal genes) are always switched on (they are constitutive) and suggest why.

32

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c


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30 Structural and Regulatory Genes other small molecules, such as microRNAs, that control the expression of structural genes. These regulatory genes may be some distance from the structural genes they control. Furthermore, within a section of DNA, expressed structural genes are enclosed on either side by untranslated regions (UTRs). UTRs contain regulatory sequences that directly control protein synthesis. How these are arranged differs between prokaryotes and eukaryotes. An overview of their structure is given below.

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Key Idea: Structural genes code for all proteins except for regulatory proteins. Regulatory genes code for the molecules involved in controlling the expression of structural genes. Genes are sections of DNA that code for proteins. They are divided broadly into structural genes and regulatory genes. Structural genes code for any protein product other than a regulatory protein. The proteins encoded by structural genes are diverse and have roles in maintaining the structure or function of a cell. Regulatory genes code for proteins and

Prokaryotic gene structure

ff In prokaryotes, several structural genes with related functions are grouped together between UTRs (below). These groupings of structural genes and their regulatory elements are called operons. The UTR upstream of the structural genes contains a regulatory sequence to initiate transcription of the structural genes. The downstream UTR stops transcription of the genes. ff In prokaryotes, the entire transcribed mRNA sequence for the structural genes is translated into proteins (there is no gene editing). Upstream of the operon, there is also a regulatory gene, which encodes a regulatory protein (not shown). Regulatory sequence

Structural genes

Regulatory sequence

DNA

UTR

Gene 1

Gene 2

Gene 3

UTR

mRNA

UTR

Gene 1

Gene 2

Gene 3

UTR

Protein 1

Protein 2

Protein 3

Eukaryotic gene structure

ff In eukaryotes, structural genes are also under the control of regulatory sequences. However, only one structural gene is enclosed by UTRs and there is no 'bulk control' of a structural gene sequence as seen in prokaryotes.

ff It is important to remember that before the primary RNA transcript is translated in eukaryotes, the non-protein coding introns are removed (only the protein-coding exons form the mature mRNA for translation).

DNA

UTR

Exon

Intron

Exon

Intron

Exon

Intron

Exon

UTR

Introns are removed leaving only the exons to be translated

mRNA

UTR

Gene 1

UTR

Protein 1

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2. How does gene expression differ in eukaryotes and prokaryotes?

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1. What is the difference between a regulatory gene and a structural gene?

LINK

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31

KNOW


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31 Eukaryotic Gene Structure and Regulation sequences ensure that transcription begins and ends at the correct points and that the necessary sequences are present to create the mature mRNA that can be exported from the nucleus. The promoter region is the binding site for regulatory proteins called transcription factors and it contains several highly conserved regions (sequences that have remained unchanged throughout evolution). Transcription's dependence on sequence recognition and transcription factors provides close control over the expression of genes.

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Key Idea: Eukaryotic genes include both translated and untranslated regions, including control sequences. Regulatory proteins called transcription factors regulate gene expression by ensuring the necessary elements come together. A eukaryotic gene includes more than just the region of exonic DNA that is translated into protein. It also contains introns, regulatory untranslated regions (UTRs), a promoter to which the RNA polymerase binds, and a terminator region that signals the stop point of transcription. The control

The structure of eukaryotic genes

A gene contains both translated and untranslated regions, including control sequences

Upstream control elements

CCAAT TATA box box

5'

DNA coding strand

Transcription start (+1) 25-35 base pairs downstream of the TATA box.

5' UTR

Enhancers

Poly-A tail signal sequence within the 3' UTR

Exon

Intron

Exon

Intron

Exon

3' UTR

Terminator sequence signals STOP transcription

Promoter region RNA polymerase recognises and binds

Highly conserved sequences within eukaryotic genes include the TATA box within the promoter and parts of the 5' and 3' UTR. These sequences must have very important functions in gene regulation.

5'

5' UTR

3'

Exon

Intron

Exon

Intron

Exon

3' UTR

3'

Primary transcript

Introns are removed to create the final mRNA

Transcribed and transported to the cytoplasm but not translated

Translation start codon (AUG)

Transcribed and removed before transport to the cytoplasm

5' cap

Transcribed and translated

Translation stop codon (UAG, UAA, or UGA) Poly A tail 3' UTR ~AAA~AAA~AAA

5' UTR

Mature mRNA

(a) Promoter region:

(b) Terminator sequence:

(c) Transcription start signal:

(d) AUG codon:

(e) UAA, UAG, and UGA codons:

2. What happens to each of the following regions of a eukaryotic gene?

(a) Exons:

(b) Introns:

(c) 5' and 3' UTR: WEB

KNOW

31

LINK

22

LINK

25

LINK

26

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1. Outline the role of each of the following regions of a eukaryotic gene:

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RNA polymerase can only transcribe genes in the presence of transcription factors ffIn eukaryotes, RNA polymerase cannot initiate

the transcription of structural genes alone. It requires the presence of transcription factors. Transcription factors are encoded by regulatory genes and have a role in creating an initiation complex for transcription.

Assembly of the transcription initiation complex Activator proteins bind to enhancer region

E n hancer

ffTranscription factors bind to distinct regions of

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the DNA, including the promoter and upstream enhancers, and act as a guide to indicate to RNA polymerase where transcription should start.

Transcription factors

ffThe TATA binding protein is a subunit of a

Coding region

multiunit general transcription factor. It is the first to bind to DNA, recruiting other transcription factors to form a transcription initiation complex. Once bound to the promoter sequence, the transcription factors capture RNA polymerase, which can then begin transcription.

Promoter region

RNA polymerase

RNA polymerase binds and transcription begins

DNA

TBP

General transcription factors including the TATA binding protein (TBP) bound to DNA

DNA hairpin loop

RNA synthesis

RNA polymerase binds to the promoter region and begins transcription

Transcription intitiation complex

3. Why would a gene contain regions that are transcribed but not translated?

4. (a) What is a transcription factor?

(b) What sort of genes encode transcription factors?

(c) How are transcription factors involved in the regulation of gene expression?

5. (a) What does it mean to say a DNA sequence is highly conserved?

(b) Some of the most highly conserved regions of genes include untranslated sequences. Why do you think this is?

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32 Models of Gene Regulation in Prokaryotes

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50

gene can produce a repressor molecule that can bind to the operator and block the transcription of the structural genes. The presence or absence of a functional repressor molecule switches the structural genes on or off and controls the metabolic pathway. Two mechanisms operate in this model: gene induction and gene repression. In gene induction (below), genes are switched on when an inducer binds to the repressor and deactivates it. In gene repression (opposite), genes are normally on but will be switched off when the endproduct of the metabolic pathway is present in excess.

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Key Idea: Prokaryotic genes are organised as operons. An operon is a cluster of several structural genes under the control of the same regulatory genes. The lac operon (lactose operon) is an operon required for the transport and metabolism of lactose in many bacteria including Escherichia coli. It is often used to study the control of gene expression in prokaryotes. The operon model applies only to prokaryotes because eukaryotic genes are not found as operons. Transcription of the structural genes in an operon is controlled by the promoter and the operator. The regulator

Prokaryotic genes occur as operons

A number of structural genes encoding the enzymes for a metabolic pathway are under the control of the same regulatory elements. Regulatory gene encodes repressor

Regulatory sequences in the untranslated region (UTR)

DNA Regulator gene

Promoter

Operator

Structural genes: translated region

Structural gene 1

Structural gene 2

Structural gene 3

OPERON

The operon consists of the structural genes and the promoter and operator sites

The regulator is some distance from the operon. It codes for the repressor that prevents the expression of specific genes.

Structural genes. At least one structural gene is present in an operon but usually there are more. The lac operon in E. coli has three. Structural genes code for the synthesis of enzymes in a metabolic pathway.

The promoter site is where the RNA polymerase enzyme first attaches itself to the DNA to begin synthesis of the mRNA.

The operator is an ‘on-off’ switch that controls RNA polymerase’s access to the structural genes. It is the repressor binding site.

The lac operon: an inducible operon

Repressor is active and the genes are normally switched off

Glucose is the preferred substrate for E. coli. The disaccharide lactose is uncommon so E. coli transcribes the genes for using it only when it is present.

An active repressor molecule binds to the operator site, switching the gene off.

RNAP

RNA polymerase cannot bind to the promoter

DNA Regulator gene

Genes are not transcribed

Promoter

Operator

lacZ

lacY

lacA

Lactose switches the lac operon on

Presence of lactose results in inactivation of the repressor so the genes can be transcribed

er

Induc

The inducer binds to the repressor altering its shape. It can no longer bind to operator site and the gene is switched on.

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When lactose is available, some of it is converted into the inducer allolactose.

Genes are transcribed

DNA Regulator gene

Promoter

Operator

lacZ

lacY

lacA

WEB

KNOW

32

CL

With the operator site free, RNA polymerase binds to the promoter and the genes for lactose metabolism are transcribed

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The trp operon: a repressible operon Transcription is normally on and must be switched off When the effector (tryptophan) is in high concentration, some binds to the trp repressor, activating it.

The combined repressor and effector bind to the operator and prevent transcription of the structural genes encoding tryptophan synthetase, the enzyme required for tryptophan synthesis.

PR E O V N IE LY W

Tryptophan repressor bound to operator.

Genes are not transcribed

Trp repressor

DNA Regulator gene

Promoter

Operator

With the operator site occupied, RNAP cannot bind to the promoter

trpE

trpD

trpC

trpB

trpA

Five structural genes encode the multiunit enzyme tryptophan synthetase

In E. coli, the enzyme tryptophan synthetase produces the amino acid tryptophan. Tryptophan is an important amino acid so the genes for producing tryptophan synthetase are normally switched on. When tryptophan is present in excess, some of it acts as an effector (or co-repressor), activating the repressor and preventing transcription of the structural genes.

1. Outline the role of each of the following components of an operon:

(a) Promoter:

(b) Operator:

(c) Structural genes:

2. What is the role of the repressor molecule in operon function?

3. Summarise the function of the lac operon by completing the following (delete the wrong answer in each choice):

(a) When lactose is absent, the repressor is active / inactive, RNA polymerase can / cannot bind, and the structural genes are / are not transcribed.

(b) When lactose is present, the repressor is active / inactive, RNA polymerase can / cannot bind, and the structural genes are / are not transcribed.

4. (a) Why is the lac operon usually switched off in E.coli?

(b) What is the advantage in having an inducible enzyme system that is regulated by the presence of a substrate?

(c) Suggest when it would not be an advantage to have an inducible system for metabolism of a substrate:

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5. How does the lac operon differ from the trp operon and explain reasons for the difference:


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33 KEY TERMS AND IDEAS: Did You Get It?

1. Match each term to its definition, as identified by its preceding letter code.

A

A gene that is involved in the production of a substance that controls or regulates the expression of one or more genes.

gene repression

B

A DNA sequence in a prokaryote operon that acts as an on-off switch when interacting with a repressor molecule.

operator

C

The increase in gene expression due to the presence of an inducer molecule.

PR E O V N IE LY W

gene induction

operon

promoter

regulatory gene structural gene TATA box

D

A gene coding for the production of a specific RNA, structural protein, or enzyme not involved in regulation.

E

The decrease in gene expression due to the activity of a repressor molecule.

F

A DNA sequence that marks the binding site of RNA polymerase.

G

A highly conserved region of the promoter to which the first general transcription factor binds prior to transcription.

H

A section of prokaryotic DNA consisting of a cluster of related structural genes, a promoter and an operator.

I

Regulatory proteins that bind to specific DNA sequences and control the transcription of DNA.

transcription factors

2. Study the diagram below, and answer the following questions. Regulatory sequence

Structural genes

UTR

Regulatory sequence

Gene 1

Gene 2

Gene 3

Protein 1

Protein 2

Protein 3

(a) What is occurring in the diagram?

(b) Is this process occurring in a eukaryote or prokaryote?

(c) Explain your decision in 2 (b):

UTR

3. Identify the regions labelled on the DNA coding strand of a eukaryotic gene below. Use the word list to help you: terminator sequence, intron, promoter, exon, transcription start signal (a)

(d) (e) 4. Use a highlighter to mark where RNA polymerase will bind. (a)

5'

TEST

(b)

(c) 5' UTR

(d)

(e)

3' UTR

CL

(c)

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(b)

3'

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Regulating biochemical pathways

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Unit 3 Outcome 1

Key terms

Enzymes are protein catalysts

activation energy

Key knowledge

active site

anabolic reaction

c

1

Outline the role of enzymes in catalysing both intracellular and extracellular biochemical reactions. Know that all the biochemical reactions in living things constitute metabolism. Distinguish between anabolic and catabolic reactions.

c

2

Explain how enzymes work to catalyse reactions, including reference to the enzyme's tertiary structure and the role of the active site, specificity, and lowering of activation energy.

c

3

Describe models for enzyme function, with reference to the enzyme-substrate complex, enzyme-product complex, and product formation.

catabolic reaction catalyst

coenzyme

denaturation enzyme

Activity number 34

34 36

35

enzyme-product complex

enzyme-substrate complex enzyme inhibition extracellular

induced fit model intracellular

Catalase

Pyruvate dehydrogenase FontanaCG cc 3.0

Restriction enzyme

irreversible inhibitor

lock-and-key model

Factors affecting enzyme activity

metabolic pathway

Key knowledge

metabolism

c

4

Describe the effect of pH, temperature, substrate concentration, and enzyme concentration on enzyme activity. Recognise that enzymes can be denatured.

37

c

5

PRAC

Investigate the effects of pH, temperature, substrate concentration, or enzyme concentration on enzyme activity.

38

c

6

Using examples, describe the effects of inhibitors on the rate of enzymecontrolled reactions. Include reference to competitive and non-competitive inhibition and identify these on graphs of reaction rate vs substrate concentration.

39

c

7

Distinguish between reversible and non-reversible inhibitors. Explain the role of reversible end-product inhibition in regulating biochemical pathways.

39

reversible inhibitor

Coenzymes in biochemical pathways Key knowledge

Activity number

c

8

Explain what is meant by a coenzyme and describe examples. Describe the general role of coenzymes in biochemical pathways.

40 41

c

9

Describe the cycling of the coenzymes ATP, NADH, and NADPH as loaded and unloaded forms move energy, protons, and electrons between cellular reactions.

40 41

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specificity

CL

optimum (for enzyme)

Activity number


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34 Enzymes

Key Idea: Enzymes are biological catalysts. The active site is critical to this functional role. Most enzymes are globular proteins. Enzymes are biological catalysts because they speed up biochemical reactions, but the enzyme itself remains unchanged. The substrate in a

reaction binds to a region of the enzyme called the active site, which is formed by the precise folding of the enzyme's amino acid chain. Enzymes control metabolic pathways. One enzyme will act on a substance to produce the next reactant in a pathway, which will be acted on by a different enzyme.

The active site

Enzymes can be intracellular or extracellular

Enzymes have an active site to which specific substrates bind. The shape and chemistry of the active site is specific to an enzyme, and is a function of the polypeptide's complex tertiary structure.

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Enzymes can be defined based on where they are produced relative to where they are active. An intracellular enzyme is an enzyme that performs its function within the cell that produces it. Most enzymes are intracellular enzymes, e.g. respiratory enzymes. Example: Catalase. Many metabolic processes produce hydrogen peroxide, which is harmful to cells. Catalase converts hydrogen peroxide into water and oxygen (below) to prevent damage to cells and tissues.

The chemical that an enzyme acts on is the substrate. An enzyme acts on a specific substrate.

2H2O + O2

2H2O2

Extremes of temperature or pH can alter the enzyme's active site and lead to loss of function. This is called denaturation.

Catalase

Substrates collide with an enzyme's active site

For a reaction to occur reactants must collide with sufficient speed and with the correct orientation. Enzymes enhance reaction rates by providing a site for reactants to come together in such a way that a reaction will occur. They do this by orientating the reactants so that the reactive regions are brought together. They may also destabilise the bonds within the reactants making it easier for a reaction to occur.

Incorrect reactant orientation = no reaction

X

Reactants

Enzyme

Enzyme orientates the reactants making reaction more likely

An extracellular enzyme is an enzyme that functions outside the cell from which it originates (i.e. it is produced in one location but active in another). Examples: Amylase and trypsin. Amylase is a digestive enzyme produced in the salivary glands and pancreas in humans. However, it acts in the mouth and small intestine respectively to hydrolyse starch into sugars. Trypsin is a protein-digesting enzyme and hydrolyses the peptide bond immediately after a basic residue (e.g. arginine). It is produced in an inactive form (called trypsinogen) and secreted into the small intestine by the pancreas. It is activated in the intestine by the enzyme enteropeptidase to form trypsin. Active trypsin can convert more trypsinogen to trypsin.

1. (a) What is meant by the active site of an enzyme and relate it to the enzyme's tertiary structure:

(b) Why are enzymes specific to one substrate (or group of closely related substrates)?

2. How do substrate molecules come into contact with an enzyme's active site?

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3. (a) Suggest why digestion (the breakdown of large macromolecules) is largely performed by extracellular enzymes:

(b) Why would an extracellular enzyme be produced and secreted in an inactive form?

WEB

KNOW

34

LINK

35

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35 Models of Enzyme Activity

Key Idea: Enzymes catalyse reactions by providing a reaction site for a substrate. The model that describes the behaviour of enzymes the best is the induced fit model. The initial model of enzyme activity was the lock and key model proposed by Emil Fischer in the 1890s. Fischer proposed enzymes were rigid structures, similar to a lock, and the substrate was the key. While some aspects of

The current induced fit model

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The lock and key model of enzyme function

Fischer's model were correct, for example, substrates align with enzymes in a way that is likely to make a reaction more likely, the model has been adapted as techniques to study molecular structures have developed. The current 'induced-fit' model of enzyme function is supported by studies of enzyme inhibitors, which show that enzymes are flexible and change shape when interacting with the substrate.

1

Enzyme

Enzyme Enzyme Enzyme

Substrate

Active site

The substrate molecule is drawn into site of the enzyme. The enzyme's active site does not change shape.

1

Enzyme Enzyme Enzyme

Substrate Substrate Substrate

Substrate Substrate Substrate

2

ES

2

3

The enzymesubstrate (ES) complex is formed.

The enzyme reaction takes place to form the enzyme-product (EP) complex.

Enzyme

Enzyme changes shape slightly Substrate as substrate binds

EP

Products Products Products

4

The products are released from the enzyme. Note there has been no change in the shape of the active site throughout the reaction.

The lock and key model proposed in 1894 suggested that the (perfectly fitting) substrate was simply drawn into a matching site on the enzyme molecule. If the substrate did not perfectly fit the active site, the reaction did not proceed. This model was supported by early X-ray crystallography studies but has since been modified to recognise the flexibility of enzymes (the induced fit model).

The enzyme changes shape as the substrate binds an enzyme-substrate (ES) complex. The shape change makes the substrate more amenable to alteration. In this way, the enzyme’s interaction with its substrate is best regarded as an induced fit.

3

The ES interaction results in an intermediate enzymeproduct (EP) complex. The substrate becomes bound to the enzyme by weak chemical bonds, straining bonds in the substrate and allowing the reaction to proceed more readily.

4

The end products are released and the enzyme returns to its previous shape.

End products released

Products

A substrate molecule is drawn into the enzyme's active site, which is like a cleft into which the substrate molecule(s) fit.

Once the substrate enters the active site, the shape of the active site changes to form an active complex. The formation of an ES complex strains substrate bonds and lowers the energy required to reach the transition state. The induced-fit model is supported by X-ray crystallography, chemical analysis, and studies of enzyme inhibitors, which show that enzymes are flexible and change shape when interacting with the substrate.

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1. Describe the key features of the ‘lock and key’ model of enzyme action and explain its deficiencies as a working model:

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2. How does the current ‘induced fit’ model of enzyme action differ from the lock and key model?

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36 How Enzymes Work

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Key Idea: Enzymes increase the rate of biological reactions by lowering the reaction's activation energy. Chemical reactions in cells are accompanied by energy changes. The amount of energy released or taken up is directly related to the tendency of a reaction to run to completion (for all the reactants to form products). Any reaction needs to raise the energy of the substrate to an unstable transition

state before the reaction will proceed (below). The amount of energy needed to do this is the activation energy (Ea). Enzymes lower the Ea by destabilising bonds in the substrate so that it is more reactive. Enzyme reactions can break down a single substrate molecule into simpler substances (catabolic reactions), or join two or more substrate molecules together (anabolic reactions).

Lowering the activation energy

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The presence of an enzyme simply makes it easier for a reaction to take place. All catalysts speed up reactions by influencing the stability of bonds in the reactants. They may also provide an alternative reaction pathway, thus lowering the activation energy (Ea) needed for a reaction to take place (see the graph below).

Without enzyme: The energy required for the reaction to proceed (Ea) is high without the enzyme present.

Transition state (unstable)

High

Amount of energy stored in the chemicals

Energy barrier

Low

Ea

High energy

Ea

With enzyme: Ea is reduced by the presence of the enzyme and the reactants form products more readily.

Reactants

Ea is the activation energy required for the reaction to begin.

Product Low energy

Start

Finish

Direction of reaction

Substrate

The substrate is attracted to the enzyme by the active site.

The substrate is cleaved (broken in two) and the two products are released to allow the enzyme to work again.

The substrate molecules are attracted to the active site.

Substrate

Product

Enzyme

Products

Stress is applied to the substrate which will help break chemical bonds.

Enzyme

Stress is applied to the substrate which will help form bonds.

The substrate molecules form a single product and are released, allowing the enzyme to work again.

Anabolic reactions

Some enzymes can cause two substrate molecules to be drawn into the active site. Chemical bonds are formed, causing the two substrate molecules to form bonds and become a single molecule. Anabolic reactions involve a net use of energy (they are endergonic) and build more complex molecules and structures from simpler ones. Examples: protein synthesis, photosynthesis.

2. Describe the difference between a catabolic and anabolic reaction:

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1. How do enzymes lower the activation energy for a reaction?

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Catabolic reactions

Some enzymes can cause a single substrate molecule to be drawn into the active site. Chemical bonds are broken, causing the substrate molecule to break apart to become two separate molecules. Catabolic reactions break down complex molecules into simpler ones and involve a net release of energy, so they are called exergonic. Examples: hydrolysis, cellular respiration.

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37 Enzyme Kinetics

Key Idea: Enzymes operate most effectively within a narrow range of conditions. The rate of enzyme-catalysed reactions is influenced by both enzyme and substrate concentration. Enzymes usually have an optimum set of conditions (e.g. of pH and temperature) under which their activity is greatest. Many plant and animal enzymes show little activity at low

temperatures. Enzyme activity increases with increasing temperature, but falls off after the optimum temperature is exceeded and the enzyme is denatured. Extremes in pH can also cause denaturation. Within their normal operating conditions, enzyme reaction rates are influenced by enzyme and substrate concentration in a predictable way.

Graph 1

Graph 2

4

Point of contact

A'

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4

A 2

B

1

0

60 C

0

3

Product formed (cm3)

Product formed (cm3)

3

120 D

Tangent

B' 2

1

0

180

0

C' 60

120 D'

Time (s)

The rate of a reaction can be calculated from the amount of product produced during a given time period. For a reaction in which the rate does not vary (graph 1) the reaction rate calculated at any one point in time will be the same. For example: B/C = A/D = A-B/D-C = (Dp/Dt) (the change in product divided by the change in time).

In a reaction in which the rate varies (graph 2) a reaction rate can be calculated for any instantaneous moment in time by using a tangent. The tangent must touch the curve at only one point. The gradient of the tangent can then be used to calculate the rate of reaction at that point in time (A'-B'/D'-C').

4

Rate of reaction

Active sites saturated

Active sites not all occupied

Product formed (cm3)

B

A

Rate of reaction

180

Time (s)

Given an unlimited amount of substrate, the rate of reaction will continue to increase as enzyme concentration increases. More enzyme means more reactions between substrates can be catalysed in any given time (graph A).

Substrate concentration

If there is unlimited substrate but the enzyme is limited, the reaction rate will increase until the enzyme is saturated, at which point the rate will remain static (graph B).

Reaction at T + 10°C

2

Reaction at T°C

0

Enzyme concentration

C

0

10

20

30

Time (s)

The effect of temperature on a reaction rate is expressed as the temperature coefficient, usually given as the Q10. Q10 expresses the increase in the rate of reaction for every rise of 10°C. Q10 = rate of reaction at (T + 10°C)/ rate of reaction at T, where T is the temperature in °C (graph C).

1. Calculate the reaction rate in graph 1:

(b) The reaction rate at 30 seconds:

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2. For graph 2: (a) The reaction rate at 90 seconds:

(b) Explain why the reaction rate in graph 2 changes over time:

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3. (a) What must be happening to the reaction mix in graph 1 to produce the straight line (constant reaction rate)?

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10

20

30

40

50

Urease

Pepsin

Antarctic icefish 1

2

3

4

5

6

7

8

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0

Trypsin

Enzyme activity

Rapid denaturation at high temperature

Too cold for the enzyme to operate

Professor Dr. habil. Uwe Kils CC3.0

Enzyme activity

Optimum temperature for enzyme

Acid

Temperature (°C)

Higher temperatures speed up all reactions, but few enzymes can tolerate temperatures higher than 50–60°C. The rate at which enzymes are denatured (change their shape and become inactive) increases with higher temperatures. The temperature at which an enzyme works at its maximum rate is called the optimum temperature.

9

10

Alkaline

pH

Enzymes performing the same function in species in different environments are very slightly different in order to maintain optimum performance. For example, the enzyme acetylcholinesterase has an optimum temperature of -2°C in the nervous system of an Antarctic icefish but an optimum temperature of 25°C in grey mullet found in the Mediterranean.

Like all proteins, enzymes are denatured by extremes of pH (very acid or alkaline). Within these extremes, most enzymes have a specific pH range for optimum activity. For example, digestive enzymes are specific to the region of the gut where they act: pepsin in the acid of the stomach and trypsin in the alkaline small intestine. Urease catalyses the hydrolysis of urea at a pH near neutral.

4. (a) Describe the change in reaction rate when the enzyme concentration is increased and the substrate is not limiting:

(b) Suggest how a cell may vary the amount of enzyme present:

5. Describe the change in reaction rate when the substrate concentration is increased (with a fixed amount of enzyme):

(b) Explain why most enzymes perform poorly at low temperatures:

(c) For graph C on the previous page, calculate the Q10 for the reaction:

7. (a) State the optimum pH for each of the enzymes: Pepsin:

Urease:

(b) Explain how the pH optima of each of these enzymes is suited to its working environment:

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Trypsin:

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6. (a) Describe what is meant by an optimum temperature for enzyme activity:

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38 Investigating Catalase Activity

Key Idea: The rate of a reaction can be measured indirectly by measuring the volume of reaction products. A group of students decided to use cubes of potato, which naturally contain the enzyme catalase, placed in hydrogen Aim To investigate the effect of potato mass (and therefore enzyme concentration) on the rate of H2O2 decomposition.

peroxide to test the effect of enzyme concentration on reaction rate. The reaction rate could be measured by the volume of oxygen produced as the hydrogen peroxide was decomposed into oxygen and water.

Timed for 5 minutes.

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Oxygen released by the reaction

Water in the 50 cm3 cylinder is displaced by the oxygen.

Hypothesis A greater mass of potato will have more enzyme present and will produce a greater reaction rate. Method The students cut raw potato into cubes with a mass of one gram. These were placed a conical flask with excess hydrogen peroxide (right). The reaction was left for five minutes and the volume of oxygen produced recorded. The students recorded the results for three replicates each of 1, 2, 3, 4, and 5 cubes of potato below:

Mass of potato (g)

Volume oxygen (cm3) (5 minutes)

Test 1

Test 2

Mean

Tube transfers released oxygen

Potato cubes + excess H2O2

A 50 cm3 cylinder is upturned in a small dish of water, excluding the air.

Mean rate of O2 production (cm3 min-1)

Test 3

1

6

5

6

2

10

9

9

3

14

15

15

4

21

20

20

5

24

23

25

1. Complete the table by filling in the mean volume of oxygen produced and the rate of oxygen production.

2. Plot the mass of the potato vs the rate of production on the grid (right):

3. Relate the rate of the reaction to the amount of enzyme present.

4. Why did the students add excess H2O2 to the reaction?

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5. State one extra reaction that should have been carried out by the students:

(b) Explain this result:

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6. (a) The students decide to cook some potato and carry out the test again with two grams of potato. Predict the result:

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39 Enzyme Inhibition

Key Idea: Enzyme activity can be reduced or stopped by inhibitors. These may be competitive or non-competitive. Enzyme activity can be stopped, temporarily or permanently, by chemicals called enzyme inhibitors. Competitive inhibitors compete directly with the substrate for the active

Competitive inhibitors compete with the normal substrate for the enzyme's active site.

site and their effect can be overcome by increasing the concentration of available substrate. A non-competitive inhibitor does not occupy the active site, but distorts it so that the substrate and enzyme can no longer interact.

Competitive inhibition

Enzyme

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Fig.1 Effect of competitive inhibition on enzyme reaction rate at different substrate concentration

Substrate

Active site

Maximum rate

Inhibitor

Substrate

Inhibitor is present in the cell (or solution) with the substrate

Inhibitor temporarily binds to the active site, blocking it so that the substrate cannot bind

Rate of reaction

Competitive inhibitors compete directly with the substrate for the active site, and their effect can be overcome by increasing the substrate concentration.

no inhibitor

competitive inhibitor

Substrate concentration

Non-competitive inhibition

Non-competitive inhibitors bind with the enzyme at a site other than the active site.

Maximum rate

Rate of reaction

They inactivate the enzyme by altering its shape so that the substrate and enzyme can no longer interact.

Fig.2 Effect of non-competitive inhibition on enzyme reaction rate at different substrate concentration

Substrate

Inhibitor

Non-competitive inhibition cannot be overcome by increasing the substrate concentration.

Enzyme

Substrate

Without the inhibitor bound, the enzyme can bind the substrate

Active site cannot bind the substrates

When the inhibitor binds, the enzyme changes shape.

no inhibitor

non-competitive inhibitor

Substrate concentration

1. Distinguish between competitive and non-competitive inhibition:

(b) Suggest how you could distinguish between competitive and non-competitive inhibition in an isolated system:

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2. (a) Compare and contrast the effect of competitive and non-competitive inhibition on the relationship between the substrate concentration and the rate of an enzyme controlled reaction (figures 1 and 2 above):

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40 Coenzymes

Key Idea: Coenzymes are needed for the function of some enzymes. They play important roles in the movement of energy, electrons, and protons about the cell. Many enzymes can only function when working with a nonprotein organic component called a coenzyme. The coenzyme can be attached to the enzyme to make it functional, e.g. by

helping the enzyme recognise the substrate. Alternatively, it may be free floating and participate so that the reaction occurs, e.g. ATP provides the energy for many biological reactions. ATP, NADH, and NADPH are three coenzymes with important roles in moving energy, protons, and electrons around the cell.

Cycling coenzymes Example of a coenzyme: ATP

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Many coenzymes work by transferring chemical groups between enzyme controlled reactions. After the chemical group has been transferred, the coenzyme must be regenerated in a separate reaction. In this way, a cycle is formed between the loaded coenzyme (ready for the reaction) and the unloaded coenzyme (needing to be regenerated).

Glucose

ATP is important in transferring energy around the cell. During an enzymecontrolled reaction, both ATP and the reactants dock with the enzyme. A phosphate group is transferred to the reactant from ATP, forming ADP and a phosphorylated reactant. The phosphorylated reactant now has the energy to react with another reactant. ADP is regenerated to ATP during cellular respiration. Thus ATP cycles as a loaded form (ATP) and an unloaded form (ADP).

NADH

NADH functions as a coenzyme in redox reactions. It is found in two forms in the cell, NAD+ and NADH. During cellular respiration (right) NAD+ accepts electrons and protons from reactants in the citric acid cycle and transfers them as NADH to the electron transport chain where they are used to provide the energy for the production of ATP.

Glucose 6 phosphate

CH2OH

CH2O

Hexokinase (enzyme)

ATP

ADP

ATP

ADP

Pyruvate

Pyruvate kinase

P

P

Phosphoenolpyruvate

In the first step of glycolysis, the enzyme hexokinase catalyses the transfer of a phosphate group from ATP to glucose, producing the more reactive glucose-6-phosphate. In the last step of glycolysis ATP is regenerated.

O-

O

Malate Odehydrogenase

NAD+

O-

NADH

O-

O

Malate

O

O

O

Oxaloacetate

NAD+

NADH

Electrons

Enzymes of the electron H+ transport chain

NADPH

NADPH is a chemically similar to NADH but has an extra phosphate group. Unlike NADH is not involved in ATP synthesis. In plants it acts as an electron carrier in photosynthesis. Like NADH it is found in two forms: NADPH and NADP+. Light is used to add electrons (and hydrogen) to NADP+ to form NADPH. NADPH then carries the electron to a second set of reactions where it is used to fix carbon.

1. (a) What is a coenyzme?

(b) What is the general function of a coenzyme?

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3. Explain how NADH is able to couple redox reactions in a cell:

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2. Explain how ATP functions as a coenzyme and is able to provide energy for cellular reactions:

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41 Achieving Metabolic Efficiency

Key Idea: Metabolic pathways are linked biochemical reactions that occur within organisms to maintain life. Metabolic reactions often occur in as a linked series in which each step in the pathway relies on the completion of a previous step and each step is controlled by specific enzymes. The end product of one enzyme-controlled step

Achieving efficiency by inhibition

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Achieving efficiency by compartmentation

provides the substrate for the next step, so failure of one step causes failure of all later steps. Metabolic pathways are tightly controlled to prevent energy being wasted. This energy conservation is termed metabolic efficiency. Metabolic reactions are often localised within specific organelles so that all the components of the pathway are kept together.

To increase metabolic efficiency, regions within a cell or an organelle are compartmentalised (separated) by membranes. Particular metabolic reactions are restricted to certain regions where all the necessary metabolic components are located. Compartmentation prevents interference between different reaction pathways and enables radically different reaction environments to be accommodated within different organelles.

Example: cellular respiration in the mitochondrion The membrane system of the mitochondrion divides it into several regions. Glycolysis takes place outside of the mitochondrion, in the cell's cytoplasm, but the remaining steps take place in different specialised regions of the mitochondrion. This helps to regulate movement of substrates and end-products and therefore reaction rates, increasing efficiency of the process (below).

Many metabolic pathways are controlled by feedback inhibition (negative feedback loop). The pathway is stopped when there is a build-up of end product (or certain intermediate products). The build-up stops the enzymes in the pathway from working and allows the cell to shut down a pathway when it is not needed. This conserves the cell's energy, so it is not manufacturing products it does not need. Both linear pathways (e.g. glycolysis), and cyclic pathways (e.g. the Krebs cycle) and can be regulated this way (below).

Glucose

Activates

Enzymes

Pyruvate

Cytoplasm (outside the mitochondrion): Glycolysis Matrix: Link reaction. Link reaction enzymes (e.g. pyruvate dehydrogenase complex) are in the matrix.

Matrix: Krebs cycle. Krebs cycle enzymes (e.g. fumarase) are in the matrix.

Matrix

Activates

Enzymes

Inhibits

Acetyl-CoA

Inhibits

ATP

Citrate Krebs cycle (6 carbon molecule)

Cristae: Electron transport chain. Membrane-bound enzymes include ATP synthase

Mitochondrion

WMU

ADP

NADH (H+)

Electron transport chain

1. What does metabolic efficiency mean?

(a) Compartmentation:

(b) Feedback inhibition:

3. What would happen if cells could not regulate their metabolic pathways?

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2. Describe how cells achieve metabolic efficiency through:

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42 KEY TERMS AND IDEAS: Did You Get It?

1. (a) What type of metabolic reaction is taking place in the diagram right?

Energy

(c) Give an example of this type of metabolic reaction:

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(b) What is occurring during this reaction?

2. Identify the following statements as true of false (circle one) (a) Enzymes are biological catalysts. They lower the activation energy of a reaction.

True / False

(b) Competitive inhibition is when an inhibitor binds to a site other than the active site.

True / False

(c) The induced fit model states that the enzyme changes shape when a substrate fits into the active site.

True / False

(d) End product inhibition causes a feedback loop that escalates the outcome of the loop.

True / False

3. The graph (right) shows the effect of an enzyme inhibitor in enzyme reaction rate. (a) It show competitive inhibition/non-competitive inhibition (delete one).

(b) Circle the diagram below that illustrates your choice in (a):

Maximum rate no inhibitor

Rate of reaction

with inhibitor

Substrate concentration

4. (a) Label the graph, right, with appropriate axes and the following labels: Reactants, products, activation energy, transition state.

(b) Assume the reaction has had no enzyme added. Draw the shape of the graph if an enzyme was added to the reaction mix.

5. Match each term to its definition, as identified by its preceding letter code.

activation energy

A Coenzyme that acts as the cell's energy carrier, transferring phosphate groups from itself to target molecules.

ATP

B A type of enzyme inhibition in which the substrate and inhibitor compete to bind to

biological catalyst

C The currently accepted model for enzyme function.

the active site.

D A globular protein that acts as a catalyst to speed up a specific biological reaction.

enzyme

E A substance or molecule that lowers the activation energy of a reaction but is

itself not used up during the reaction. In biological systems this function is carried out by enzymes.

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competitive inhibition

F The energy required for a reactant to reach an unstable transition state in which it can react with another reactant.

G The compound on which an enzyme acts. non-competitive inhibition

H A type of enzyme inhibition in which the inhibitor does not occupy the active site but binds to some other part of the enzyme.

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induced fit model

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Photosynthesis

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Unit 3 Outcome 1

Key terms

The role of photosynthesis

absorption spectrum

Key knowledge

accessory pigment action spectrum ATP

Activity number

c

1

Recall the role of photosynthesis in supporting the majority of life on Earth.

c

2

Write the overall equation for photosynthesis in words and symbols. What is the role of light in the process?

Calvin cycle carotenoid

Chloroplasts are the site of photosynthesis

chloroplast

Key knowledge

chlorophyll

43

43 48

Activity number

44 46 48

c

3

Describe the structure and role of chloroplasts, including the thylakoid membranes and liquid stroma. Explain the role of chlorophyll a and b, and accessory pigments (e.g. carotenoids) in light capture.

light dependent phase

c

4

46

light independent phase

Distinguish between the absorption spectrum and action spectrum with respect to light absorbing pigments.

c

5

PRAC

47

NADP/NADPH

c

6

Describe the evidence for the bacterial origin of chloroplasts.

chromatography grana

Use thin layer chromatography to separate photosynthetic pigments.

45

photolysis

photosynthesis photosystem stroma

stroma lamellae thylakoid discs

triose phosphate

Kristian Peters

Stages in photosynthesis

Dartmouth College

Activity number

Key knowledge

7

Describe the inputs and outputs of the light dependent stage of photosynthesis.

48

c

8

Outline the events in the light dependent reactions of photosynthesis, including absorption of light, transfer of excited electrons between carriers in the thylakoid membranes, generation of ATP and NADPH, and the photolysis of water.

48

c

9

PRAC

Investigate the effect of a named factor, e.g. light intensity, on the rate of dehydrogenase activity in extracts of chloroplasts.

50

c

10

Describe the inputs and outputs of the light independent stage of photosynthesis (the Calvin cycle), including the role of the enzyme RuBisCo.

48

c

11

Outline the events in the Calvin cycle, including the fixation of carbon and the production of triose phosphate using reduced NADPH and ATP.

48

c

12

Outline the evidence for the processes occurring in the light dependent and light independent reactions of photosynthesis.

49

Factors affecting photosynthesis Key knowledge

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c

Activity number

13

Describe and explain limiting factors for photosynthesis (to include carbon dioxide concentration, light intensity, and temperature).

51

c

14

PRAC

Investigate the effect of environmental factors, e.g. light intensity, on the rate of photosynthesis.

53

c

15

Outline methods used to overcome limiting factors for photosynthesis, including the use of controlled environments.

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43 The Role of Photosynthesis

Key Idea: Photosynthesis is the chemical process in which autotrophs use sunlight to produce carbohydrates. Photosynthesis is the process by which plants capture light energy and use it to fix (convert) the carbon in CO2 into carbohydrates (e.g. glucose). The carbohydrate is used by

the plant to power ATP production and build its body. Plants (and other photosynthetic organisms) carry out this process without input from other organisms, so they are called producers (as opposed to consumers, which depend on energy and carbon from other organisms).

Photosynthesis and producers ff A producer (or autotroph) is an organism that

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can make its own food.

Carbon dioxide gas (CO2)

Sugar (stored energy) and water

ff Plants, algae, and some bacteria are producers. ff Most producers use the energy in sunlight to

Water

make their food. The process by which they do this is called photosynthesis. Photosynthesis transforms sunlight energy into chemical energy.

Sunlight

ff The chemical energy is stored as glucose,

and the energy is released when the glucose undergoes further metabolic processes.

ff The inputs and outputs of photosynthesis are shown on the leaf diagram (right).

Photosynthesis by marine algae provides oxygen and absorbs carbon dioxide. Most algae are microscopic but some, like this kelp, are large.

Depending on the plant, 0.1% to 8% of the light intercepted is used in photosynthesis. Typically crop plants use about 1%-2%.

Oxygen gas (O2)

USDA

The evolution of oxygenic photosynthesis, in which water is split to provide the hydrogens to drive the process, was responsible for our current oxygen rich atmosphere. Using water (rather than hydrogen sulfide) to supply hydrogens provided far more energy for ATP production and produced oxygen gas as a waste product.

Producers, such as grasses, make their own food, and are also the ultimate source of food and energy for consumers, such as these cattle.

1. (a) What is a producer?

2. Where do producers get their energy from?

3. Why are producers so important in an ecosystem?

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(b) Name some organisms that are producers:

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44 Chloroplasts

Key Idea: Chloroplasts have a complicated internal membrane structure. They are the site of photosynthesis in plant cells. Chloroplasts are the specialised plastids in which photosynthesis occurs. A mesophyll (photosynthetic) leaf cell contains between 50-100 chloroplasts. The chloroplasts are generally aligned so that their broad surface runs parallel to the cell wall to maximise the surface area available for

light absorption. Chloroplasts have an internal structure characterised by a system of membranous structures called thylakoids arranged into stacks called grana. Special pigments, called chlorophylls and carotenoids, are bound to the membranes as part of light-capturing photosystems. They absorb light of specific wavelengths and thereby capture the light energy.

The structure of a chloroplast

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Thylakoid membranes provide a large surface area for light absorption. They are the site of the light dependent phase and are organised so as not to shade each other.

Chloroplast is enclosed by a double membrane envelope (inner and outer membrane)

Liquid stroma contains the enzymes for the light independent phase. It also contains the chloroplast's DNA.

Chloroplasts

Starch granule

Lipid droplet

Stroma lamellae connect the grana. They account for 20% of the thylakoid membrane.

TEM image of a single chloroplast

Kristian Peters

Grana (sing. granum) are stacks of thylakoids

Cell wall

Chloroplasts visible in plant cells

1. Label the transmission electron microscope image of a chloroplast below:

(a)

(d)

(b)

(e)

(c) (c)

(f)

Image: Dartmouth College

2. (a) Where is chlorophyll found in a chloroplast?

(b) Why is chlorophyll found there?

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4. Explain why plant leaves appear green:

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3. Explain how the internal structure of chloroplasts helps absorb the maximum amount of light:

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45 The Bacterial Origins of Chloroplasts

Key Idea: Chloroplasts arose as an endosymbiosis between an early pre-eukaryotic cell and a cyanobacterium. Oxygenic photosynthesis is thought to have evolved in cyanobacteria about 2.3 billion years ago (bya) and eukaryotes evolved about 1 bya. It is thought that eukaryotic cells evolved from pre-eukaryotic (bacterial) cells that ingested other freefloating bacteria. They formed a symbiotic relationship with

the cells they engulfed (endosymbiosis). The two organelles that evolved in eukaryotic cells as a result of bacterial endosymbiosis were mitochondria, for aerobic respiration, and chloroplasts, for photosynthesis in aerobic conditions. Primitive eukaryotes probably acquired mitochondria by engulfing purple bacteria. Similarly, chloroplasts may have been acquired by engulfing photosynthetic cyanobacteria.

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Evolution of a plant cell

Original preeukaryotic cell

Oxygen-using bacterium engulfed

Photosynthetic cyanobacterium engulfed

Plant cell

Chloroplast

Animals

Mitochondrion

Examples of engulfment

Evidence for the bacterial origin of chloroplasts includes: `` Chloroplasts have a similar morphology to bacteria.

Paramecium bursaria Paramecium bursaria (right) is a single celled protozoan. It engulfs cells of Zoochlorella, a photosynthetic green alga. It houses the algae and carries them to light areas in a pond where they can photosynthesise. In return, it uses the food made by the algae.

`` Chloroplasts divide by binary fission, they split in half to form new chloroplasts, just like bacteria. Thus new chloroplasts arise from preexisting chloroplasts, they are not manufactured by the cell.

`` They have a chemically distinct inner membrane. The outer

membrane is similar to the plasma membrane (as if a vesicle formed around the engulfed bacteria) but the inner membrane is similar to the membrane of bacteria.

`` Bacterial DNA is a single circular molecule and chloroplasts also have their own single circular DNA. Like bacterial DNA, the DNA of chloroplasts has no introns or histones. Also the chloroplast DNA evolves or mutates at a different rate to the nuclear DNA.

`` Chloroplasts contain ribosomes that are more similar in size to bacterial ribosomes than ribosomes in the cytoplasm.

`` Antibiotics that inhibit protein synthesis in bacteria also inhibit

protein synthesis in chloroplasts. Conversely, bacterial toxins that inhibit protein synthesis in eukaryotes do not affect chloroplasts.

`` Analysis of chloroplast DNA has shown that they are related to cyanobacteria.

Bob Blaylock

Evidence for the bacterial origin of chloroplasts

Amoeba proteus From 1972 microbiologist Kwang Jeon studied the infection of Amoeba proteus by Legionella-like bacteria. He found that most infected amoebae died. The few that survived were cultured over many generations. Eventually, the amoebae became dependent on the bacteria for nuclear function. Experiments showed that when the nucleus of an infected cell was placed in an uninfected cell which had also had its nucleus removed the new cell quickly died.

1. Which endosymbiosis occurred first in the evolution of eukaryotic cells? Explain your reasoning:

2. Outline four pieces of evidence for the bacterial origin of chloroplasts: (a)

(c) (d)

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(b)

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3. How do the examples of Paramecium bursaria and Amoeba proteus support the endosymbiotic origin of chloroplasts?

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46 Pigments and Light Absorption

Key Idea: Chlorophyll pigments absorb light of specific wavelengths and capture light energy for photosynthesis. Substances that absorb visible light are called pigments, and different pigments absorb light of different wavelengths. The ability of a pigment to absorb particular wavelengths of light can be measured with a spectrophotometer. The light absorption vs the wavelength is called the absorption

spectrum of that pigment. The absorption spectrum of different photosynthetic pigments provides clues to their role in photosynthesis, since light can only perform work if it is absorbed. An action spectrum profiles the effectiveness of different wavelengths of light in fuelling photosynthesis. It is obtained by plotting wavelength against a measure of photosynthetic rate (e.g. O2 production).

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The electromagnetic spectrum

10-5 nm

10-3 nm

Gamma rays

103 nm

1 nm

Ultra violet

X- rays

106 nm

Infrared

103 m

1m

Microwaves

Radio waves

Visible light

380

450

550

Increasing energy

650

750

Increasing wavelength

Wavelength (nm)

100

EMR travels in waves, where wavelength provides a guide to the energy of the photons. The greater the wavelength of EMR, the lower the energy of the photons in that radiation.

The photosynthetic pigments of plants

Absorption spectra of photosynthetic pigments

(Relative amounts of light absorbed at different wavelengths)

80

Chlorophyll b

60

Carotenoids

The photosynthetic pigments of plants fall into two categories: chlorophylls (which absorb red and blue-violet light) and carotenoids (which absorb strongly in the blueviolet and appear orange, yellow, or red). The pigments are located on the chloroplast membranes (the thylakoids) and are associated with membrane transport systems.

Chlorophyll a

40

Green light reflected

Chloroplast

Su

Absorbance (percent)

Light is a form of energy known as electromagnetic radiation (EMR). The segment of the electromagnetic spectrum most important to life is the narrow band between about 380 nm and 750 nm. This radiation is known as visible light because it is detected as colours by the human eye. It is visible light that drives photosynthesis.

t

gh

nli

20

Action spectrum for photosynthesis

(Effectiveness of different wavelengths in fuelling photosynthesis)

100

Red and blue light absorbed

Thylakoid discs

80

The pigments of chloroplasts in higher plants (above) absorb blue and red light, and the leaves therefore appear green (which is reflected). Each photosynthetic pigment has its own characteristic absorption spectrum (top left). Only chlorophyll a participates directly in the light reactions of photosynthesis, but the accessory pigments (chlorophyll b and carotenoids) can absorb wavelengths of light that chlorophyll a cannot and pass the energy (photons) to chlorophyll a, thus broadening the spectrum that can effectively drive photosynthesis.

60

40

20

0 400

The action spectrum and the absorption spectrum for the photosynthetic pigments (combined) match closely. 500

600

700

Wavelength (nm)

1. What is meant by the absorption spectrum of a pigment?

Left: Graphs comparing absorption spectra of photosynthetic pigments compared with the action spectrum for photosynthesis.

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Rate of photosynthesis (as % of rate at 670 nm)

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2. Why doesn't the action spectrum for photosynthesis exactly match the absorption spectrum of chlorophyll a?

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47 Separation of Pigments by Chromatography

Key Idea: Photosynthetic pigments can be separated from a mixture using chromatography. Chromatography involves passing a mixture dissolved in a mobile phase (a solvent) through a stationary phase, which separates the molecules according to their

specific characteristics (e.g. size or charge). In thin layer chromatography, the stationary phase is a thin layer of adsorbent material (e.g. silica gel or cellulose) attached to a solid plate. A sample is placed near the bottom of the plate which is placed in an appropriate solvent (the mobile phase).

Separation of photosynthetic pigments

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The four primary pigments of green plants can be easily separated and identified using thin layer chromatography. The pigments from the leaves are first extracted by crushing leaves, together with acetone, using a mortar and pestle. The extract is dotted on to the chromatography plate. Acetone is used as the mobile phase (solvent). During thin layer chromatography, the pigments separate out according to differences in their relative solubilities. Two major classes of pigments are detected: the two greenish chlorophyll pigments and two yellowish carotenoid pigments.

Determining Rf values

Cling wrap or tin foil to seal beaker

Leaf extract

Original solvent front

Solvent

Solvent front A

1. (a) Calculate the Rf values for the pigments A-D on the chromatography plate shown left.

(b) Use the Rf values to identify the pigments:

A: Rf value:

Pigment: B: Rf value:

B

C

Pigment: C: Rf value:

Pigment:

D

D: Rf value:

Pigment:

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2. A student carried out a chromatography experiment in class. The instructions said to leave the plate in the solvent for 30 minutes, but the student instead removed the plate after 20 minutes. How would this affect the Rf values and pigment separations obtained?

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Sample application point

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48 Photosynthesis: Inputs and Outputs

Key Idea: Photosynthesis is the process by which light energy is used to convert CO2 and water into glucose and oxygen. Photosynthesis is of fundamental importance to living things because it transforms sunlight energy into chemical energy stored in molecules, releases free oxygen gas, and absorbs carbon dioxide (a waste product of cellular metabolism). Photosynthesis has two phases, the light dependent phase and the light independent phase. In the reactions of the light

dependent phase, light energy is converted to chemical energy (ATP and NADPH). This phase occurs in the thylakoid membranes of the chloroplasts. In the reactions of the light independent phase, the chemical energy is used to synthesise carbohydrate. This phase occurs in the stroma of chloroplasts. In photosynthesis, water is split and electrons are transferred together with hydrogen ions from water to CO2, reducing it to triose phosphates (then converted to sugars).

Light independent phase (LIP):

In the first phase of photosynthesis, chlorophyll captures light energy, which is used to split water, producing O2 gas (waste). Electrons and H+ ions are transferred to the molecule NADPH. ATP is also produced. The light dependent phase occurs in the thylakoid membranes of the grana.

The second phase of photosynthesis occurs in the stroma and uses the NADPH and the ATP to drive a series of enzyme-controlled reactions (the Calvin cycle) that fix carbon dioxide to produce triose phosphate. This phase does not need light to proceed.

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Light dependent phase (LDP):

Sun

Chloroplast outer membrane

ligh

t

LDP

H2O (waste)

LIP

O2

ATP

LDP

CO2

Converted to

NADPH

Triose phosphate (C3H7O6P)

Grana

Thylakoid

Rubisco is the central enzyme in the LIP of photosynthesis (carbon fixation) catalysing the first step in the Calvin cycle. However it is remarkably inefficient, processing just three reactions a second. To compensate, rubisco makes up almost half the protein content of chloroplasts.

CO2 from the air provides raw materials for glucose production.

Diagrammatic representation of a chloroplast

Monosaccharides (e.g. glucose) and other carbohydrates, lipids, and amino acids.

The general equation for photosynthesis

6CO2 + 6H2O

Light

Chlorophyll

C6H12O6 + 6O2

1. Identify the two phases of photosynthesis and their location in the cell: (a) (b)

2. (a) What is the role of the enzyme Rubisco?

3. State the origin and fate of the following molecules involved in photosynthesis:

(a) Carbon dioxide:

(c) Hydrogen:

(b) Oxygen:

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(b) Rubisco is the most abundant protein on Earth. Suggest a reason for this:

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4. Use the information on the opposite page to fill in the diagram below, including the raw material (inputs), products (outputs), and processes.

Raw materials

(a)

(c)

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ADP

ATP

Light dependent phase

Solar energy

(f) Process:

Light independent phase (h) Process:

(g) Location:

Main product

(e)

(i) Location:

NADPH

NADP+

(b)

By-products

(d)

5. In two experiments, radioactively-labelled oxygen (shown in blue) was used to follow oxygen through the photosynthetic process. The results of the experiment are shown below:

Experiment A: 6CO2 + 12H2O + sunlight energy C6H12O6 + 6O2 + 6H2O

Experiment B: 6CO2 + 12H2O + sunlight energy C6H12O6 + 6O2 + 6H2O

From these results, what would you conclude about the source of the oxygen in:

(a) The carbohydrate produced?

(b) The oxygen released?

6. Name the products that triose phosphate is converted into:

(a) The light dependent phase of photosynthesis:

(b) The light independent phase of photosynthesis:

8. What is the function of each of the following in photosynthesis:

(a) ATP:

(b) NADPH:

(c) Light:

(d) Chlorophyll:

(e) Water:

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7. Describe what happens during:


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49 How Do We Know: Photosynthesis

Key Idea: Hill's experiment using isolated chloroplasts and Calvin's "lollipop" experiment provided important information on the process of photosynthesis. In the 1930s Robert Hill devised a way of measuring

oxygen evolution and the rate of photosynthesis in isolated chloroplasts. During the 1950s Melvin Calvin led a team using radioisotopes of carbon to work out the steps of the light independent reactions (the Calvin cycle).

Robert Hill's experiment

Calvin's lollipop experiment Melvin Calvin and his colleges placed the algae Chlorella vulgaris in a thin bulb shaped flask to simulate a leaf (the lollipop).

Radioactive 14C labelled CO2 was bubbled into the flask at precise times.

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The dye DCPIP (2,6-dichlorophenol-indophenol) is blue. It is reduced by H+ ions and forms DCPIPH2 (colourless). Hill made use of this dye to show that O2 is produced during photosynthesis even when CO2 is not present.

Light

1000

0

Power

Speed

RPM

2000

BBT Inc.

Speed

Timer

On/Off

Leaves are homogenised to form a slurry. The slurry is filtered to remove any debris. The filtered extract is then centrifuged at low speed to remove the larger cell debris and then at high speed to separate out the chloroplasts.

Samples were taken two seconds apart and were placed directly into boiling ethanol which stops any reactions.

The samples were analysed to work out which molecules the 14C was incorporated into.

Two dimensional chromatography was used to separate the molecules in each sample. The sample is run in one direction using a solvent, then turned 90° and run again using a second solvent. This separates molecules that are similar in size (below). Second solvent front

Dark

Light

First solvent front

The chloroplasts are resuspended in a buffer. The blue dye DCPIP is added to the suspension. In a test tube left in the dark, the dye remains unchanged. In a test tube exposed to the light, the blue dye fades and the test tube turns green again. The rate of colour change can be measured by measuring the light absorbance of the suspension. The rate is proportional to the rate at which oxygen is produced.

Hill's experiment showed that water must be the source of oxygen (and therefore electrons). It is split by light to produce H+ ions (which reduce DCPIP) and O2- ions. The equation below summarises his findings: H2O + A → AH2 + ½ O2

where A is the electron acceptor (in vivo this is NADP+)

Direction of first run

Direction of second run

Original sample

The chromatography plate was placed on X-ray film to find the position of the molecule(s) with the radioactive 14C. The molecule could be isolated from the chromatography plate and analysed to find its structure. By taking samples a few seconds apart it was possible to work out the order in which the molecules carrying the 14C appeared and therefore the steps involved in fixing CO2 into glucose.

2. What important finding about photosynthesis did Hill's experiment show?

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1. Write an equation for the formation of DCPIPH2 from DCPIP:

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3. Why did the samples in Calvin's lollipop experiment need to be taken just seconds apart?

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50 Investigating Enzymes in Photosynthesis

Key Idea: Replacing NADP+ with DCPIP as the electron acceptor allows the effect of light on the rate of the light dependent reactions to be measured indirectly. NADP+ is the electron acceptor for the light dependent

reaction. By substituting the dye DCPIP, which fades from blue to colourless as it accepts electrons, it is possible to indirectly measure the rate of the light dependent reactions and therefore the rate of enzyme activity during the reactions.

Background

Results Tube number (absorbance) Time (min)

1

2

3

4

5

0

5.0

5.0

5.0

0.3

5.0

1

4.8

5.0

-

0.3

5.0

2

4.7

4.9

-

0.3

5.0

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Dehydrogenase enzymes play a role in the transport of electrons through the photosystem pathways of the light dependent reactions. The final acceptor of the electron is NADP+, forming NADPH. By substituting DCPIP to accept H+, the rate of enzyme activity can be measured.

The aim

To investigate the effect of light intensity on the rate of dehydrogenase activity in the light dependent reactions in isolated chloroplasts.

3

4.6

4.8

-

0.3

5.0

4

4.3

4.8

-

0.4

5.0

The method

5

4.0

4.7

-

0.3

4.9

Pieces of spinach leaf were blended using a standard food processor. The pulp was filtered through a muslin cloth into four centrifuge tubes kept in an ice bath. The filtered extract was spun down to produce a pellet (solid) and supernatant (liquid). The supernatant was discarded and each pellet resuspended with a medium of cold sucrose solution in a boiling tube. In tube 1 and 2 the dye DCPIP was added. In tube 3 DCPIP was added then the tube was covered in foil to exclude light. In tube 4 no DCPIP was added. In a fifth tube DCPIP and sucrose medium were added without any leaf extract. Tubes 1 and 3 were exposed to high intensity light. Tube 2 was exposed to a lower intensity light. The absorbance of all the tubes was measured using a colorimeter at time 0 and every minute for 15 minutes. The colorimeter records the fading of the blue colour. The absorbance of tube 3 was measured at the beginning and end of the experiment only.

6

3.8

4.6

-

0.4

4.9

7

3.4

4.6

-

0.2

4.9

8

3.0

4.5

-

0.3

5.0

9

2.6

4.4

-

0.4

5.1

10

2.2

4.4

-

0.3

5.0

11

1.9

4.3

-

0.2

4.9

12

1.4

4.1

-

0.2

5.0

13

0.9

4.0

-

0.3

4.8

14

0.6

4.0

-

0.3

5.0

15

0.5

3.8

4.7

0.4

5.0

1. Write a brief hypothesis for this experiment:

2. Use the grid (right) to draw a line graph of the change in absorbance over time of each of the tubes tested. 3. (a) What was the purpose of tube 4?

(b) What was the purpose of tube 5?

6. Write a conclusion for the investigation:

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5. Why did the absorbance of tubes 4 and 5 vary?

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4. Why was the absorbance of tube 3 only measured at the start and end of the investigation?

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Key Idea: Environmental factors, such as light availability and carbon dioxide level, affect photosynthetic rate. The rate at which plants can photosynthesise is dependent on environmental factors, particularly light and carbon dioxide (CO2), but also temperature. In the plant's natural environment, fluctuations in these factors (and others)

influence photosynthetic rate, so that the rate varies daily and seasonally. The effect of each factor can be tested experimentally by altering one while holding the others constant. Usually, either light or CO2 level is limiting. Humans can overcome the limitations of low light or CO2 by growing plants in a controlled environment.

Factors Affecting Photosynthetic Rate 90

Figure A: Effect of light intensity on photosynthetic rate

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The rate of photosynthesis is affected by abiotic (non-living) factors in the environment, particularly light intensity, temperature, and CO2 level. The two graphs (right) illustrate the effect of these variables on the rate of photosynthesis in cucumber plants. Figure A shows the effect of increasing light intensity at constant temperature and CO2 level. Figure B illustrates how this response is influenced by CO2 concentration and temperature. 30°C is at the upper range of tolerance for many plants.

Rate of photosynthesis (mm3 CO2 cm–2 h–1)

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51 Factors Affecting Photosynthesis

80

70

60

50

40

1

Plants acquire CO2 from the atmosphere through their stomata. They also lose water through their stomata when they are open, so conditions that cause them to close their stomata to reduce water loss also reduce CO2 uptake and lower the photosynthetic rate. Such conditions include increased wind speed and water stress. Glasshouses are used to manipulate the physical environment to maximise photosynthetic rates.

Rate of photosynthesis (mm3 CO2 cm–2 h–1)

Glasshouse environments can artificially boost CO2 levels

280

2

3

4

5

6

7

Figure B: Effect of light intensity, temperature, and CO2 on High CO2 at 30°C photosynthetic rate

240 200

High CO2 at 20°C

160 120

Low CO2 at 30°C

80

Low CO2 at 20°C

40

1

High wind increases water loss from stomata

2

3

4

5

6

Units of light intensity (arbitrary scale)

7

(a) CO2 concentration:

(b) Light intensity:

(c) Temperature:

2. Why does photosynthetic rate decline when the CO2 level is reduced?

3. Why might hot, windy conditions reduce photosynthetic rates?

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1. Based on the figures above, summarise and explain the effect of each of the following factors on photosynthetic rate:

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52 Overcoming Limiting Factors in Photosynthesis environment system such as a greenhouse. Temperature, carbon dioxide (CO2) concentration, and light intensity are optimised to maximise the rate of photosynthesis and therefore production. Greenhouses also allow specific abiotic factors to be manipulated to trigger certain life cycle events such as flowering. CO2 enrichment dramatically increases the growth of greenhouse crops providing that other important abiotic factors (such as nutrients) are not limiting.

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Key Idea: Growing plants in controlled environments allows manipulation of abiotic factors to maximise yield. Manipulating abiotic factors can maximise crop yields by maximising photosynthetic rate and reducing losses to pests, disease, and competition. For example, covering the soil with black plastic reduces weed growth, increases soil temperature, and boosts production. Greater control of the abiotic conditions is achieved by growing crops in a controlled-

Large, commercial greenhouses have elaborate computer-controlled watering systems linked to sensors that measure soil moisture, air temperature, and humidity. Coupled with a timer, they deliver optimal water conditions for plant growth by automatically operating the irrigation system.

Carbon dioxide enrichment ``Carbon dioxide (CO2) is a raw material used in

photosynthesis. If the supply of CO2 is cut off or reduced, plant growth and development are curtailed. The amount of CO2 in air is normally 0.03% (250-330 ppm). Most plants will stop growing when the CO2 level falls below 150 ppm. Even at 220 ppm, a slowing of plant growth is noticeable (see graph, right).

``Controlled CO2 atmospheres, which boost the CO2

concentration to more than 1000 ppm, significantly increase the rate of formation of dry plant matter and total yield (e.g. of flowers or fruit). Extra carbon dioxide can be generated (at a cost) by burning hydrocarbon fuels, using compressed, bottled CO2 or dry ice, or by fermentation or decomposition of organic matter.

Air flow through a greenhouse is essential to providing a homogeneous air temperature. Air flow also ensures an even distribution of CO2 throughout the enclosure. Airflow from one end of the enclosure to the other is maintained by fans all blowing in the same direction.

The effect of CO2 concentration on plant growth

Percentage of normal growth rate

The growing environment can be controlled or modified to varying degrees. Black plastic sheeting can be laid over the soil to control weeds and absorb extra solar heat. Tunnel enclosures (above) may be used to reduce light intensity and airflow, prevent frost damage, and reduce damage by pests.

250%

200%

150%

Normal growth rate: 100%

100% 50% 0%

0

200

400

600 800 1000 1200 1400 1600 1800

300 (normal air)

No growth

Carbon dioxide (ppm) Enriched levels

Toxic

1. Explain why CO2 enrichment has the capacity to radically increase crop production:

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3. List the abiotic factors that are controlled in a greenhouse environment:

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2. State the two primary considerations influencing the economic viability of controlled environments:

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53 Investigating Photosynthetic Rate

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Key Idea: Measuring the production of oxygen provides a simple means of measuring the rate of photosynthesis. The rate of photosynthesis can be investigated by measuring the substances involved in photosynthesis. These include

The aim

To investigate the effect of light intensity on the rate of photosynthesis in an aquatic plant, Cabomba aquatica.

measuring the uptake of carbon dioxide, the production of oxygen, or the change in biomass over time. Measuring the rate of oxygen production provides a good approximation of the photosynthetic rate and is relatively easy to carry out. Cabomba aquatica, a common aquarium plant

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The method `` 0.8-1.0 grams of Cabomba stem were weighed on a balance. The stem was cut and inverted to ensure a free flow of oxygen bubbles.

`` The stem was placed into a beaker filled with a solution

containing 0.2 mol L-1 sodium hydrogen carbonate (to supply carbon dioxide). The solution was at approximately 20°C. A funnel was inverted over the Cabomba and a test tube filled with the sodium hydrogen carbonate solution was inverted on top to collect any gas produced.

Piotr Kuczynski CC 3.0

76

`` The beaker was placed at distances (20, 25, 30, 35, 40,

45, 50 cm) from a 60W light source and the light intensity measured with a lux meter at each interval.

`` Before recording data, the Cabomba stem was left to

acclimatise to the new light level for 5 minutes. Because the volumes of oxygen gas produced are very low, bubbles were counted for a period of three minutes at each distance.

The results

Light intensity (lx) (distance)

Bubbles counted Bubbles per in three minutes minute

5 (50 cm)

0

13 (45 cm)

6

30 (40 cm)

9

60 (35 cm)

12

95 (30 cm)

18

150 (25 cm)

33

190 (20 cm)

35

1. Complete the table by calculating the rate of oxygen production (bubbles of oxygen gas per minute): 2. Use the data to draw a graph of the bubble produced per minute vs light intensity:

3. Although the light source was placed set distances from the Cabomba stem, light intensity in lux was recorded at each distance rather than distance per se. Explain why this would be more accurate:

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4. The sample of gas collected during the experiment was tested with a glowing splint. The splint reignited when placed in the gas. What does this confirm about the gas produced?

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5. What could be a more accurate way of measuring the gas produced in the experiment?

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54 KEY TERMS AND IDEAS: Did You Get It?

1. (a) Write the process of photosynthesis as: A word equation: A chemical equation:

(b) Where does photosynthesis occur?

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2. Test your vocabulary by matching each term to its correct definition, as identified by its preceding letter code. absorption spectrum accessory pigments

A The biochemical process that uses light energy to convert carbon dioxide and water into glucose molecules and oxygen.

B The stacks of thylakoids within the chloroplasts of plants.

action spectrum

C Membrane-bound compartments in chloroplasts. They are the site of the light

Calvin cycle

D The phase in photosynthesis in which chemical energy is used for the synthesis of

dependent reactions of photosynthesis.

carbohydrate. Also called the light independent phase.

chlorophyll

E The liquid interior of the chloroplast where the light independent phase takes place.

grana

F The phase in photosynthesis when light energy is converted to chemical energy.

light dependent phase

G The term to describe the light absorption of a pigment vs the wavelength of light.

photosynthesis

H Plant pigments that absorb wavelengths of light that chlorophyll a does not absorb. I A profile of the effectiveness of different wavelengths of light in fuelling

stroma

photosynthesis.

thylakoid discs

J The green, membrane-bound pigment involved in the light dependent reactions of photosynthesis.

3. Label the following features of a chloroplast on the diagram below: granum, stroma, thylakoid disc, stroma lamellae

4. In the image below part of a variegated leaf is covered with aluminium foil. On the outline of the leaf below draw or shade the parts of the leaf where you would expect photosynthesis to occur:

Green

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White

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Foil

TEST


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Cellular respiration

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Unit 3 Outcome 1

Key terms

The role of cellular respiration

acetyl coA

Key knowledge

alcoholic fermentation

anaerobic metabolism ATP

c

1

Recall the role of ATP in the energy transformations of living systems. Describe the purpose of cellular respiration in living organisms.

c

2

Write the overall equation for cellular respiration in words and symbols.

ATP synthase

cellular respiration

Glycolysis

cristae

Key knowledge

decarboxylation

Activity number

c

3

dehydrogenation

electron transport chain

55 56 58

Activity number

Outline the location and inputs and outputs of glycolysis, including ATP yield. Recognise glycolysis as the first step in cellular respiration and an almost universal pathway in cells.

58

ethanol FAD

fermentation glucose

glycolysis

Krebs cycle

lactic acid fermentation

Dartmouth College

Masur

link reaction

Mitochondria are the site of cellular respiration

matrix

Key knowledge

mitochondrion

EII

Activity number 57 58

c

4

Describe the structure and role of mitochondria, including the cristae (folded inner membranes) and matrix.

oxidative phosphorylation

c

5

Describe the evidence for the bacterial origin of mitochondria. Why do scientists think that mitochondria evolved before chloroplasts in the evolution of eukaryotic cells?

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pyruvate

c

6

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substrate level phosphorylation

Describe the inputs and outputs of the Krebs cycle and electron transport chain, including the role of redox reactions and electron carriers, and the ATP yield.

c

7

PRAC

Use a simple respirometer to investigate respiration rate, e.g. in germinating seeds, under different conditions.

59

NAD/NADH

Anaerobic pathways

Activity number

Key knowledge

8

Describe the location, inputs, and outputs of anaerobic pathways for ATP generation in eukaryotes to include alcoholic fermentation in yeast and lactic acid fermentation in mammalian muscle.

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9

Compare and explain ATP yield from fermentation and aerobic respiration.

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c

Factors affecting cellular respiration Key knowledge

Activity number

58 60 62

10

Describe factors affecting the rate of cellular respiration, including temperature, glucose availability, and oxygen concentration.

c

11

PRAC

Investigate the effect of temperature on aerobic respiration in yeast. Yeast respire aerobically when sugars are limited. Can you explain why?

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c

12

PRAC

Investigate the effect of substrate type on rate of fermentation in yeast.

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55 The Role of ATP in Cells

Key Idea: ATP transports chemical energy within the cell for use in metabolic processes. All organisms require energy to be able to perform the metabolic processes required for them to function and reproduce. This energy is obtained by cellular respiration, a set of metabolic reactions which ultimately convert

biochemical energy from 'food' into the nucleotide adenosine triphosphate (ATP). ATP is considered to be a universal energy carrier, transferring chemical energy within the cell for use in metabolic processes such as biosynthesis, cell division, cell signalling, thermoregulation, cell mobility, and active transport of substances across membranes.

Adenosine triphosphate (ATP)

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The ATP molecule consists of three components; a purine base (adenine), a pentose sugar (ribose), and three phosphate groups which attach to the 5' carbon of the pentose sugar. The structure of ATP is shown right.

Adenine

Phosphate groups

The bonds between the phosphate groups contain electrons in a high energy state which store a large amount of energy. The energy is released during ATP hydrolysis. Typically, hydrolysis is coupled to another cellular reaction to which the energy is transferred. The end products of the reaction are adenosine diphosphate (ADP) and an inorganic phosphate (Pi). Note that energy is released during the formation of bonds during the hydrolysis reaction, not the breaking of bonds between the phosphates (which requires energy input).

Ribose

ATP powers metabolism

Solid particle

The energy released from the removal of a phosphate group of ATP is used for active transport of molecules and substances across the plasma membrane e.g. phagocytosis (above) and other active transport processes.

Chromosomes

Mitosis, as observed in this onion cell, requires ATP to proceed. Formation of the mitotic spindle and chromosome separation both require energy from ATP hydrolysis to occur.

CDC

CDC

Mitotic spindle

ATP is required when bacteria divide by binary fission (above). For example, ATP is required in DNA replication and to synthesise components of the peptidoglycan cell wall.

Not all of the energy released in the oxidation of glucose is captured in ATP. The rest is lost as heat. This heat energy can be used to maintain body temperature. Thermoregulatory mechanisms such as shivering and sweating also use ATP.

1. What process produces ATP in a cell?

2. On the space filling model of ATP shown right, label adenine, ribose and the phosphate groups:

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4. Describe one other process in a cell that requires ATP:

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3. Explain why thermoregulation requires the expenditure of energy:

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56 ATP and Energy

Key Idea: ATP is the universal energy carrier in cells. Energy is stored in the covalent bonds between phosphate groups. The molecule ATP (adenosine triphosphate) is the universal energy carrier for the cell. ATP can release its energy quickly by hydrolysis of the terminal phosphate. This reaction is

catalysed by the enzyme ATPase. Once ATP has released its energy, it becomes ADP (adenosine diphosphate), a low energy molecule that can be recharged by adding a phosphate. The energy to do this is supplied by the controlled breakdown of glucose in cellular respiration.

How does ATP provide energy? The reaction of A + B is endergonic. It requires energy to proceed and will not occur spontaneously.

PR E O V N IE LY W

ATP releases its energy during hydrolysis. Water is split and added to the terminal phosphate group resulting in ADP and Pi. For every mole of ATP hydrolysed 30.7 kJ of energy is released. Note that energy is released during the formation of chemical bonds not from the breaking of chemical bonds.

Adenosine

Mitochondrion

ATP is reformed during the reactions of cellular respiration (i.e. glycolysis, Krebs cycle, and the electron transport chain).

P

P

A

B

P

Hydrolysis is the addition of water. ATP hydrolysis gives ADP + Pi (HPO42-) + H+.

The enzyme ATPase is able to couple the hydrolysis of ATP directly to the formation of a phosphorylated intermediate A-Pi

ATPase

Adenosine

P

P

A

+

Pi

A-Pi is more reactive than A. It is now able to react with B.

Inorganic phosphate

Pi

+

A

B

A-Pi reacts with B and Pi is released.

In reality these reactions occur virtually simultaneously.

In many textbooks the reaction series above is simplified and the Note! The phosphate bonds in ATP are often referred to as high energy bonds. This can be misleading. The bonds contain intermediates are left out: electrons in a high energy state (making the bonds themselves AB A+B relatively weak). A small amount of energy is required to break the bonds, but when the intermediates recombine and form ATP ADP + Pi new chemical bonds a large amount of energy is released. The final product is less reactive than the original reactants.

(b) In what way is the ADP/ATP system like a rechargeable battery?

2. What respiratory substrate provides the energy for reforming ATP?

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1. (a) How does ATP supply energy to power metabolism?

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3. During the many metabolic reactions occurring in the body, most of the energy in the initial respiratory substrate is lost as heat. What is the purpose of this heat?

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57 Mitochondrial Structure and Bacterial Origin subsequent similar event introduced chloroplasts into some eukaryotic cells, creating the lineage that led to plants. Many of the lines of evidence for chloroplasts being of bacterial origin are the same for the origin of mitochondria. These include mitochondria having circular DNA, smaller ribosomes, and independent binary fission. Cellular respiration and ATP production occur in mitochondria. The inner and outer mitochondrial membranes create distinct regions where the different reactions of cellular respiration take place.

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Key Idea: Mitochondria evolved after an endosymbiosis between a pre-eukaryote and an aerobic bacterial cell. Their structure helps compartmentalise enzyme reactions. As in the origin of chloroplasts, the evidence indicates that the mitochondria of eukaryotic cells arose as a result of an endosymbiotic relationship between a precursor eukaryotic cell and a bacterial cell. The endosymbiotic event that generated mitochondria must have happened early in the history of eukaryotes, because all eukaryotes have them. A

The mitochondrion

Amine oxidases on the outer membrane surface Phosphorylases between the inner and outer membranes

ATPases on the inner membranes (the cristae)

WMU

Soluble enzymes for the Krebs cycle and fatty acid degradation floating in the matrix

Drawing of mitochondrion

TEM Transverse section of a mitochondrion

Evolution of a eukaryotic cell

Infolding membrane

Developing nucleus

Mitochondrion

Original preeukaryotic cell

Internal membranes including the endoplasmic reticulum are formed by infoldings of the plasma membrane.

Aerobically respiring purple bacteria engulfed.

Eukaryotic cell

1. (a) Describe the general role of mitochondria in cell respiration:

(b) Explain the importance of compartmentation in the mitochondrion:

2. Explain how mitochondria evolved:

(b) (c) 4. How do we know mitochondria evolved before chloroplasts?

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(a)

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3. Identify three pieces of evidence that mitochondria evolved from engulfed bacteria:

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58 Cellular Respiration: Inputs and Outputs

Key Idea: During cellular respiration, the energy in glucose is transferred to ATP in a series of enzyme controlled steps. The oxidation of glucose is a catabolic, energy yielding pathway. The breakdown of glucose and other organic fuels to simpler molecules is coupled to ATP synthesis. Glycolysis and the Krebs cycle supply electrons to the electron transport chain (ETC), which drives oxidative phosphorylation. The

conversion of pyruvate (the end product of glycolysis) to acetyl CoA links glycolysis to the Krebs cycle. Most of the ATP generated in cellular respiration is produced by oxidative phosphorylation when NADH + H+ and FADH2 donate electrons to the electron carriers in the ETC. At the end of the chain, electrons are passed to molecular oxygen, reducing it to water. Electron transport is coupled to ATP synthesis.

Overview of cellular respiration Mitochondrion outer membrane

PR E O V N IE LY W

Respiration involves three metabolic stages (plus a link reaction) summarised below. The first two stages are the catabolic pathways that decompose glucose and other organic fuels. In the third stage, the electron transport chain accepts electrons from the first two stages and passes these from one electron acceptor to another. The energy released at each stepwise transfer is used to make ATP.

Cristae: folded inner membrane. Enzymes for the electron transport chain reside here

Matrix: enzymes for the Krebs cycle reside here.

Glucose

NADH

Glycolysis

Pyruvate

Link reaction

2 ATP

tyl Ace me y z n coe A

Krebs cycle

NADH FADH 2

Electron transport chain

2 ATP

34 ATP

The theoretical maximum yield of 38 ATP per mole of glucose has recently been revised down to 32 ATP (28 from the ETC). All figures will be updated in subsequent revisions.

The general equation for cellular respiration

C6H12O6 + 6O2

6CO2 + 6H2O + energy

1. Describe precisely in which part of the cell the following take place:

(a) Glycolysis:

(b) The link reaction: (c) Krebs cycle reactions:

(d) Electron transport chain:

2. Write a word equation for the general equation for cellular respiration:

5. Describe two functions of the Krebs cycle in the process of cellular respiration:

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4. Describe three functions of glycolysis in the process of cellular respiration:

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3. What is the total number of ATP produced from one glucose molecule?

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Steps in cellular respiration Glycolysis

Oxidative phosphorylation

An enzyme transfers a phosphate group directly from a substrate (such as glucose) to ADP to form ATP. Net ATP yield from substrate level phosphorylation below.

Glucose is oxidised in a series of reduction and oxidation (redox) reactions that provide the energy for the formation of ATP. Net ATP yield from oxidative phosphorylation below.

Glycolysis Glucose 2ATP

2NADH

6ATP

2NADH

6ATP

6NADH

18ATP

2FADH2

4ATP

PR E O V N IE LY W

Glycolysis is the beginning of cellular respiration. It takes glucose and produces two pyruvate molecules, each of which can then enter the Krebs cycle. Glycolysis initially uses two ATP but produces four ATP. NADH is produced for use in the electron transport chain. The numbers shown are for one glucose molecule.

Substrate level phosphorylation

Link reaction

Pyruvate

The link reaction removes CO2 from pyruvate and adds coenzyme A, producing the 2C molecule acetyl coenzyme A, which enters the Krebs cycle. NADH is also produced for use in the electron transport chain.

Pyruvate

In the Krebs cycle, acetyl coenzyme A is attached to the 4C molecule oxaloacetate and coenzyme A is released. Oxaloacetate is eventually remade in a cyclic series of reactions that produce more NADH and FADH2 for the electron transport chain. Two ATP are also made by substrate level phosphorylation.

Acetyl coenzyme A

2CO2

Krebs cycle

2ATP

Krebs cycle

Electron transport chain

Electrons carried by NADH and FADH2 are passed to a series of electron carrier enzymes embedded in the inner membrane of the mitochondria. The energy from the electrons is used to pump H+ ions across the inner membrane from the matrix into the intermembrane space. These are allowed to flow back to the matrix via the enzyme ATP synthase which uses their energy to produce ATP. The electrons are coupled to H+ and oxygen at the end of the electron transport chain to form water.

4CO2

Total

Electron transport chain

H+

H+

H+

H+

NADH

34ATP

e–

NAD+

FADH2

FAD

H2O

H+

H+ + ½O

2

6. (a) What is substrate level phosphorylation?

(b) How many ATP are produced this way during cellular respiration (per molecule of glucose)?

7. (a) What is oxidative phosphorylation?

(b) How many ATP are produced this way during cellular respiration (per molecule of glucose)?

10. What is the purpose of NADH and FADH2 in cellular respiration?

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9. Describe how ATP is produced in the electron transport chain:

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8. Which parts of cellular respiration produce CO2?


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59 Measuring Respiration

Key Idea: Oxygen consumption and carbon dioxide production in respiring organisms can be measured with a respirometer. A respirometer measures the amount of oxygen consumed Screw clip

Scale

Measuring respiration with a simple respirometer

Coloured bubble

The diagram on the left shows a simple respirometer. It measures the change in gases as respiration occurs. ff Respiring organisms, in this case germinating seeds, are placed into the bottom of the chamber.

PR E O V N IE LY W

Capillary tube

and the amount of carbon dioxide produced during cellular respiration. Respirometers are quite simple pieces of apparatus but can give accurate results if set up carefully.

ff Soda lime or potassium hydroxide is added to absorb any carbon dioxide produced during respiration. Therefore the respirometer measures oxygen consumption.

ff Once the organisms have been placed into the chamber the screw clip is closed. The start position of the coloured bubble is measured (this is the time zero reading).

ff The coloured bubble in the capillary tube moves in response to

Perforated metal cage Germinating seeds

the change in oxygen consumption. Measuring the movement of the liquid (e.g. with a ruler) allows the change in volume of gas to be estimated.

ff Care needs to be taken when using a simple respirometer because changes in temperature or atmospheric pressure may change the readings and give a false measure of respiration.

ff Differential respirometers (not shown) use two chambers

Soda lime (or KOH) pellets (CO2 absorbant)

Caution is required when handling KOH as it is caustic. Wear protective eyewear and gloves.

(a control chamber with no organisms and a test chamber) connected by a U-tube. Changes in temperature or atmospheric pressure act equally on both chambers. Observed changes are only due to the activities of the respiring organism.

1. Why does the bubble in the capillary tube move?

Clamp stand

2. A student used a simple respirometer (like the one above) to measure respiration in maggots. Their results are presented in the table (right). The maggots were left to acclimatise for 10 minutes before the experiment was started.

(a) Calculate the rate of respiration and record this in the table. The first two calculations have been done for you.

(b) Plot the rate of respiration on the grid, below right.

0

0

_

5

25

5

10

65

15

95

20

130

25

160

(d) Why was there an acclimatisation period before the experiment began?

3. Why would it have been better to use a differential respirometer?

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(c) Describe the results in your plot:

Rate (mm min-1)

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Distance bubble moved (mm)

Time (minutes)

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60 Investigating Aerobic Respiration in Yeast

Key Idea: Respiration rate can be investigated using the redox indicator triphenyl tetrazolium chloride (TTC). In simple organisms such as yeast, respiration rate can be affected by different factors, including temperature and availability of respiratory substrates. Yeast is a facultative anaerobe and will respire aerobically when glucose levels

are low and oxygen is available. The effect of temperature on yeast respiration rate can be determined using the redox indicator triphenyl tetrazolium chloride (TTC). TTC is a colourless hydrogen acceptor and intercepts the hydrogen ions produced during respiration. It turns red when reduced and the rate of colour change indicates the rate of respiration.

Background

TTC indicator (1 mL)

Yeast suspension (10 mL)

PR E O V N IE LY W

During respiration, dehydrogenase enzymes remove hydrogens from glucose and pass them to hydrogen acceptors. TTC intercepts these hydrogens and turns red. The rate of colour change indicates the rate of enzyme activity and thus the rate of respiration.

Aim and hypothesis

To investigate the effect of temperature on the rate of aerobic respiration in yeast. If respiration is occurring, the rate can be indicated by the amount of colour change in the indicator TTC.

The method

A set of three test tubes (tubes 1-3) were set up containing 10 mL of dilute yeast suspension (10 g yeast, 1.5 g glucose per litre) and 1 mL of TTC. At this concentration of glucose and in the presence of oxygen, yeast will respire aerobically.

The test tubes were placed into a water bath at 25 °C. Two more sets of three test tubes were prepared in the same way. The second set was placed into a water bath at 40 °C (tubes 4-6), and the third set was placed into a water bath at 55°C (tubes 7-9). The rate of colour change was measured over 4.5 hours by measuring the absorbance of each tube with a colorimeter (0 being clear or low absorbance and 10 being fully opaque or high absorbance). A control tube containing only yeast and glucose was included in the experiment (tube 10).

Water bath 25 °C

Caution needs to be taken with TTC use as it can cause skin irritation. Gloves should be used.

Tube number (absorbance)

25°C

40°C

55°C

Control

Time (hr)

1

2

3

4

5

6

7

8

9

10

1.5

1.13

1.02

1.20

2.34

2.33

2.29

4.11

4.05

4.17

0.40

3.0

1.96

1.88

2.04

5.85

5.89

5.80

8.86

8.90

8.82

0.51

4.5

2.76

2.69

2.81

7.84

7.88

7.80

9.77

9. 87

9.74

0.62

1. Calculate the mean absorbance for each of the times and temperatures above and enter the values in the table below: Mean absorbance

Time (hr)

25°C

40°C

55°C

1.5 3.0 4.5

2. Use the table in (1) to plot a graph of absorbance over time for the three temperatures measured:

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4. How does temperature affect the respiration rate of yeast?

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3. Why did the absorbance of the control tube change?

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61 Anaerobic Pathways

Key Idea: Glucose can be metabolised aerobically and anaerobically to produce ATP. The ATP yield from aerobic processes is higher than from anaerobic processes. Aerobic respiration occurs in the presence of oxygen. Organisms can also generate ATP when oxygen is absent by

using a molecule other than oxygen as the terminal electron acceptor for the pathway. In alcoholic fermentation in yeasts, the electron acceptor is ethanal. In lactic acid fermentation, which occurs in mammalian muscle even when oxygen is present, the electron acceptor is pyruvate itself.

Alcoholic fermentation

Lactic acid fermentation

In alcoholic fermentation, the H+ acceptor is ethanal which is reduced to ethanol with the release of carbon dioxide (CO2). Yeasts respire aerobically when oxygen is available but can use alcoholic fermentation when it is not. At ethanol levels above 12-15%, the ethanol produced by alcoholic fermentation is toxic and this limits their ability to use this pathway indefinitely. The root cells of plants also use fermentation as a pathway when oxygen is unavailable but the ethanol must be converted back to respiratory intermediates and respired aerobically.

Glucose

Skeletal muscles produce ATP in the absence of oxygen using lactic acid fermentation. In this pathway, pyruvate is reduced to lactic acid, which dissociates to form lactate and H+. The conversion of pyruvate to lactate is reversible and this pathway operates alongside the aerobic system all the time to enable greater intensity and duration of activity. Lactate can be metabolised in the muscle itself or it can enter the circulation and be taken up by the liver to replenish carbohydrate stores. This 'lactate shuttle' is an important mechanism for balancing the distribution of substrates and waste products.

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Glucose 2 ADP

2 ADP

2 ATP net

2 ATP net

NADH + H+

NADH + H+

2 x pyruvate

2 x pyruvate

CH3COCOOH

CH3COCOOH

Lactic acid fermentation

Yeast, higher plant cells

Animal tissues

NAD+

Waste product

NADH + H+

Ethanal

Pyruvate

CH3CHO

CH3COCOOH

Gaseous waste product

The alcohol and CO2 produced from alcoholic fermentation form the basis of the brewing and baking industries. In baking, the dough is left to ferment and the yeast metabolises sugars to produce ethanol and CO2. The CO2 causes the dough to rise.

+ CO2

+

NADH + H+

Yeasts are used to produce almost all alcoholic beverages (e.g. wine and beers). The yeast used in the process breaks down the sugars into ethanol (alcohol) and CO2. The alcohol produced is a metabolic byproduct of fermentation by the yeast.

Lactate

CH3 CHOHCOO– + H+

+

NAD+

Wintec

C6H12O6

Alcoholic fermentation

Ethanol

CH3CH2OH

C6H12O6

The lactate shuttle in vertebrate skeletal muscle works alongside the aerobic system to enable maximal muscle activity. Lactate moves from its site of production to regions within and outside the muscle (e.g. liver) where it can be respired aerobically.

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1. Describe the key difference between aerobic respiration and fermentation:

(b) Why is the efficiency of these anaerobic pathways so low?

3. Why can't alcoholic fermentation go on indefinitely?

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2. (a) Refer to pages 82-83 and determine the efficiency of fermentation compared to aerobic respiration:

%

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62 Investigating Fermentation in Yeast

Key Idea: Brewer's yeast preferentially uses alcoholic fermentation when there is excess sugar, releasing CO2, which can be collected as a measure of fermentation rate. Brewer's yeast is a facultative anaerobe (meaning it can respire aerobically or use fermentation). It will preferentially

use alcoholic fermentation when sugars are in excess. One would expect glucose to be the preferred substrate, as it is the starting molecule in cellular respiration, but brewer's yeast is capable of utilising a variety of sugars, including disaccharides, which can be broken down into single units.

The aim

5 minutes between readings

PR E O V N IE LY W

To investigate the suitability of different monoand disaccharide sugars as substrates for alcoholic fermentation in yeast.

Water in the 100 cm3 cylinder is displaced by the CO2.

CO2 released by the yeast fermentation

The hypothesis

Tube transfers released CO2

If glucose is the preferred substrate for fermentation in yeast, then the rate of fermentation will be highest when the yeast is grown on glucose rather than on other sugars.

Background

The rate at which brewer's or baker's yeast (Saccharomyces cerevisiae) metabolises carbohydrate substrates is influenced by factors such as temperature, solution pH, and type of carbohydrate available. The literature describes yeast metabolism as optimal in warm, acid (pH 4-6) environments.

High levels of sugars suppress aerobic respiration in yeast, so yeast will preferentially use the fermentation pathway in the presence of excess substrate.

Substrate

A 100 cm3 cylinder is upturned in a small dish of water, excluding the air.

10 g substrate + 225 cm3 water + 25 cm3 yeast culture

The apparatus

In this experiment, all substrates tested used the same source culture of 30 g active yeast dissolved in 150 cm3 of room temperature (24°C) tap water. 25 g of each substrate to be tested was added to 225 cm3 room temperature (24°C) tap water buffered to pH 4.5. Then 25 cm3 of source culture was added to the test solution. The control contained yeast solution but no substrate.

The substrates

Glucose is a monosaccharide, maltose (glucose-glucose), sucrose (glucosefructose), and lactose (glucose-galactose) are disaccharides.

1. Write the equation for the fermentation of glucose by yeast:

Volume of carbon dioxide collected (cm3)

None

Glucose

Maltose

Sucrose

Lactose

0

0

0

0

0

0

5

0

0

0.8

0

0

10

0

0

0.8

0

0

2. The results are presented on the table left. Using the final values, calculate the rate of CO2 production per minute for each substrate:

15

0

0

0.8

0.1

0

(a) None:

20

0

0.5

2.0

0.8

0

25

0

1.2

3.0

1.8

0

(b) Glucose:

30

0

2.8

3.6

3.0

0

(c) Maltose:

35

0

4.2

5.4

4.8

0

(d) Sucrose:

40

0

4.6

5.6

4.8

0

45

0

7.4

8.0

7.2

0

(e) Lactose:

50

0

10.8

8.9

7.6

0

55

0

13.6

9.6

7.7

0

60

0

16.1

10.4

9.6

0

65

0

22.0

12.1

10.2

0

70

0

23.8

14.4

12.0

0

75

0

26.7

15.2

12.6

0

80

0

32.5

17.3

14.3

0

85

0

37.0

18.7

14.9

0

90

0

39.9

21.6

17.2

0

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Experimental design and results adapted from Tom Schuster, Rosalie Van Zyl, & Harold Coller , California State University Northridge 2005

3. What assumptions are being made in this experimental design and do you think they were reasonable?

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Time (min)

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4. Use the tabulated data to plot an appropriate graph of the results on the grid provided:

5. (a) Identify the independent variable:

(b) State the range of values for the independent variable: 

(c) Name the unit for the independent variable:

6. (a) Identify the dependent variable:

(b) Name the unit for the dependent variable:

7. (a) Summarise the results of the fermentation experiment:

(b) Why do you think CO2 production was highest when glucose was the substrate?

(c) Suggest why fermentation rates were lower on maltose and sucrose than on glucose:

(d) Suggest why there may have been no fermentation on the lactose substrate:

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8. Predict what would happen to CO2 production rates if the yeast cells were respiring aerobically:

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63 KEY TERMS AND IDEAS: Did You Get It?

1. Match each term to its definition, as identified by its preceding letter code. aerobic respiration

alcoholic fermentation ATP

A A series of reactions that converts glucose into pyruvate. The energy released is used to produce ATP.

B The process in cellular respiration which involves the oxidation of glucose by a series of redox reactions that provide the energy for the formation of ATP.

C Organelles responsible for producing the cell’s ATP. They appear oval in shape with an outer double membrane and a convoluted interior membrane.

PR E O V N IE LY W

cellular respiration

electron transport chain glycolysis

D The stage in cellular respiration where pyruvate enters the mitochondrion and carbon dioxide is removed.

E An anaerobic pathway where ethanal acts as the electron acceptor and the end product is ethanol.

F A type of metabolic reaction that results in the formation of ATP by direct transfer of a phosphate group to ADP from a phosphorylated reactive intermediate.

Krebs cycle

lactic acid fermentation

G An anaerobic pathway occurring in the skeletal muscle of mammals. Pyruvate is reduced to lactic acid.

H Respiration requiring oxygen as the terminal electron acceptor.

I A nucleoside triphosphate used in the transfer of energy in cells.

link reaction

J Also known as the citric acid cycle. Part of a metabolic pathway involved in the chemical conversion of carbohydrates, fats and proteins to CO2 and water to generate a form of usable energy (ATP).

matrix

mitochondria

K The catabolic process in which the chemical energy in complex organic molecules is coupled to ATP production.

oxidative phosphorylation

L A product of glycolysis. An important intermediate in many metabolic pathways. M The region of the mitochondrion enclosed by the inner mitochondrial membrane.

pyruvate

N Chain of enzyme-based redox reactions, which passes electrons from high to low redox potentials. The energy released is used to pump protons across a membrane and produce ATP.

substrate level phosphorylation

2. Complete the diagram of cellular respiration below by filling in the boxes below:

No. ATP

(a)

No. ATP

(b)

No. ATP

(c)

(d)

CO2

(g)

3. Study the graph of oxygen consumption of lobsters, right.

(a) What happens to the rate of oxygen consumption as oxygen concentration increases?

(b) What happens to the rate of oxygen consumption as temperature increases?

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Waste

Waste

(h)

Oxygen uptake of lobster

18°C 16.5°C

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Waste

15°C

7.5°C

Oxygen concentration (mL O2 per L water)

CL

(f)

Oxygen consumption (L O2 kg-1 h-1)

Input

(e)

Link reaction

Glycolysis

TEST


64 Review: Unit 3, Area of Study 1

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90

Summarise what you know about this topic under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts. Use the images and hints to help you and refer back to the introduction to check the points covered:

Nucleic acids

HINT: Describe the relationship between DNA and proteins. Summarise DNA structure, replication, and the basics of gene expression.

Plasma membranes

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HINT: Outline the relationship between membrane structure and transport mechanisms.

Gene structure and regulation

REVISE

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HINT: Gene structure and regulation. Lac operon

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91 Regulating biochemical pathways

Photosynthesis

HINT: Chloroplast structure, function, and origin. Inputs and outputs of photosynthesis and factors affecting photosynthesis.

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HINT: The activity and regulation of enzymes.

Cellular respiration

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HINT: Mitochondrion structure, function, and origin. Cellular respiration and fermentation.


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65 Synoptic Assessment: Unit 3, Area of Study 1

1. DNA replication occurs prior to mitosis and cell division. It produces new DNA by semi-conservative replication. (a) What is the purpose of DNA replication?

(b) What does semi-conservative mean?

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2. The diagram below shows a simplified overview of DNA replication. Summarise the process by labelling the diagram according to the questions below:

(a) Label the 5' and 3' ends of all strands on the DNA being replicated.

(b) Label: (i) A parent strand (ii) A daughter strand (iii) Free nucleotides (iv) The new chromatids (v) The leading strand (vi) The lagging strand

A

(c) Draw in the positions of: (i) DNA polymerase (ii) Helicase (d) What is happening at position A?

(e) Circle the nucleotide to be added next to the leading strand and use an arrow to show where it will go.

(f) Circle the nucleotide to be added next to the lagging strand and use an arrow to show where it will go.

3. Enzymes are essential for catalysing biological reactions. (a) Describe the induced fit model of enzyme action:

(b) Draw a diagram in the space below to show how a metabolic pathway can be regulated by feedback inhibition:

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4. Aerobic cellular respiration is essential for providing usable energy for cells. Outline this process in eukaryotes (you may use extra paper if required). Your discussion should include reference to the purpose of aerobic cellular respiration, the location of the main stages, and the raw material(s) and final product(s):

5. Photosynthesis is a process by which autotrophs such as green plants use light to produce carbohydrate molecules. Outline the process of photosynthesis in plants and explain how its rate is affected by specific environmental factors (you may use extra paper if required). Your discussion should include reference to the location of photosynthesis in plant cells, the raw material(s) and final product(s), and factors that affect photosynthetic rate.

6. The diagrams below depict the lac operon in E. coli. The first structural gene is identified for reference.

(a) Label the promoter, operator, and regulator regions and identify the remaining two structural genes lacY and lacA:

(b) Rule a line to indicate the boundaries of the operon

(c) Which gene produces the repressor molecule?

DNA

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(d) Draw the position of the repressor molecule and RNA polymerase when lactose is absent. Are the structural genes transcribed?

lacZ

(e) Draw the position of the repressor molecule and RNA polymerase when lactose is present. Are the structural genes transcribed?

DNA

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lacZ

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Cellular signals

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Unit 3 Outcome 2

Key terms

Signalling molecules and target cells

apoptosis

Key knowledge

cancer

caspases cytokine

c

1

Explain what is meant by a signal molecule (or ligand) and, in a general way, explain the effect of signal molecules on target cells.

c

2

Describe the sources and modes of transmission of signalling molecules to their target cells to include hormones in plant and animals, neurotransmitters, cytokines (e.g. interferon, interleukin, and growth factors), and pheromones.

extracellular receptor extrinsic pathway

Activity number 66

66 67

first messenger

G-protein coupled receptor hormone

hydrophilic signal

hydrophobic signal

intracellular receptor intrinsic pathway

EII

Jeevan Jose cc 4.0

mitochondrial pathway neurotransmitter

The stimulus-response model and signal transduction

pheromone

3

Describe the stimulus-response model with respect to cell signalling to include reception, transduction, and cellular response.

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c

4

Distinguish between signal transduction involving hydrophilic signals (e.g. adrenaline) and hydrophobic signals (e.g. steroid hormones). Include reference to differences in how the signal molecule is received by receptors and how transduction is initiated.

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c

5

For hydrophilic signal molecules, recognise the role of G protein-coupled receptors, second messengers, and phosphorylation cascades in producing the cellular response (names of molecules not required).

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c

6

For hydrophobic signal molecules, recognise the role of nuclear receptors (transcription factors) which are activated when the signal molecule binds.

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proto-oncogenes signal molecule

signal transduction target cell

transcription factor tumour

tumour suppressor genes

Apoptosis

Activity number

Key knowledge

c

7

Describe apoptosis (programmed cell death) as an example of a cellular response to specific signals. Explain the role of apoptosis during development and in the life of an adult organism.

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c

8

Distinguish between apoptosis in response to signals from inside the cell (via the mitochondrial pathway) and outside the cell (the death receptor pathway). Explain the role of caspases (proteolytic enzymes) in these pathways.

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c

9

Describe malfunctions in apoptosis that result in deviant cell behaviour leading to disease. These malfunctions are related to upregulation or downregulation of caspase activity. Examples include:

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second messenger

Activity number

c

• Cancer and autoimmune diseases when there is too little apoptosis • Neurodegeneration, e.g. Alzheimer's, when there is too much apoptosis.

CL

phosphorylation cascade

Key knowledge


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66 What is Cell Signalling? the cell and bring about a response. Cells that possess the receptors to bind a particular signal molecule are called target cells for that signal molecule. If a cell does not have the specific receptor, then it is unaffected by the chemical signal. Chemical signals can be classified based on how far they travel to cause an effect. Some signals have a localised effect, while others can influence cells (or organisms) a considerable distance away.

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Key Idea: Cells use signals (chemical messengers) to communicate and to gather information about, and respond to, changes in their cellular environment. In order to communicate and respond to changes in their environment, cells must be able to send, receive, and process signals. Most signals are chemicals (signal molecules or ligands) and many different types exist. In order for a signal to have an effect on a cell it must be able to bind to

Endocrine signalling A signal is carried in the bloodstream to target cells, often a considerable distance away.

Endocrine cell

Target cells

Endocrine signalling may involve hormones (from endocrine cells) or cytokines as the signalling molecule.

Blood vessel

Example: The production of insulin from the pancreas stimulates the cellular uptake of glucose.

Signal molecule

Receptor

Insulin

Synapse

Paracrine signalling Some cell signalling occurs at a more local level (between cells that are close together). The signal molecule binds to the receptors of a nearby cell causing a response. Neurotransmitters or cytokines are the signalling molecules involved in this type of cellular signalling.

Presynaptic cell

Postsynaptic cell

Signal molecule (neurotransmitter)

Synapse

Example: Neurotransmitters released from a nerve cell travel across the synapse (gap) to another cell to cause a response.

T cell

Signal molecules

Autocrine signalling Cells can produce and react to their own signals. This type of signalling is important during growth and development and in the functioning of the immune system. Example: In vertebrates, the presence of a foreign antibody causes T-cells to produce a growth factor to stimulate their own production. The increased number of T-cells helps to fight the infection.

Receptor

B

Pheromone Binding Protein (PBP) with bound bomykol (B)

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Pheromones Some organisms produce chemical signals that are secreted externally to cause a response in other organisms. The signal molecules are called pheromones, and they affect the physiology or behaviour of members of the same species.

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Pheromone (bombykol) secreted by the female silk moth (Bombyx mori) is detected by the male's antennae causing his wings to flutter.

PBP

Example: The pheromone bombykol is released by female silk moths and is detected in very low concentrations by males. The pheromone binding protein on the male's antennae and binds the bombykol, triggering a wing fluttering response. He uses the chemical concentration gradient to locate his mate.

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What is a receptor?

In order for a chemical signal to have an effect on a cell it must bind to a receptor. Cell receptors are protein molecules, and are very specific about the type of chemical signal they bind. Each receptor is linked to a specific biochemical pathway. Once the chemical signal has been bound a message is passed to the target cell causing a cellular response. The specificity of the receptor or a particular signal molecule prevents cells wasting resources and energy on producing cellular products that are not needed.

The binding sites of cell receptors are specific only to certain ligands (signal molecules). This stops them reacting to every signal the cell encounters. Receptors generally fall into two main categories:

Extracellular domain

Outside of cell

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Intracellular receptors

Extracellular receptors

Extracellular receptors (also called transmembrane receptors) span the cell membrane and bind ligands that cannot cross the plasma membrane. They have an extracellular domain outside the cell, and an intracellular domain within the cell cytosol. A diagram of an extracellular receptor is shown on the right.

Cell membrane

Intracellular receptors (also called cytoplasmic receptors) are located within the cell cytoplasm and they bind ligands that are able to cross the plasma membrane.

Cell cytosol

Intracellular domain

Structure of a transmembrane receptor

1. Briefly describe the types of cell signalling:

2. Identify the components that all signalling types have in common:

3. (a) Why doesn't every cell respond to a signal molecule?

(b) Why is it important the cells don't respond to every signal molecule produced by an organism?

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Cellular response

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4. In the space below label the sequence (1-3) in the correct order:

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67 The Nature of Signalling Molecules as hormones) tend to be slow acting and long lasting, while others (such as neurotransmitters) take effect rapidly, but the effect is short lived as the chemical is quickly broken down. Most signals affect cells within the organism, but pheromones are secreted into the environment and may travel over long distances to influence members of the same species.

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Key Idea: Signalling molecules are widespread in nature. The effect they have is highly varied and they may act over short or long distances to cause a response. Signalling molecules bind to specific receptors to cause a response in a target cell. There are a huge number of different signal molecules, and each has a specific effect. Some (such

Gibberellins break dormancy in seeds

Plant hormones (phytohormones) have important roles in plant growth and development (e.g. stem elongation, breaking dormancy, and fruit fall). Plant hormones are transported around the plant by the plant's vascular tissue. They often work together and the response varies depending on the relative concentrations of each.

A nerve cell

Neurotransmitters are chemicals that carry signals between nerve cells or between a nerve cell and another type of cell such as a muscle or gland. Neurotransmitters act on the cell immediately next to it. They are released into a synapse (gap between the cells) and bind to receptors on the receiving cell. The response is rapid and short lived.

Hormones help animals prepare for and adjust to seasonal changes

In mammals, hormones are secreted by endocrine glands (e.g. the pituitary) and carried in the blood to target cells. Hormones are very potent, and effective at low concentrations. Hormonal responses tend to be slow (because it takes time for the signal to reach its target) and generally long lasting because they induce metabolic changes.

Lymphocytes produce cytokines

Cytokines are a large group of peptides and small proteins involved in coordinating the response of cells in the immune system both within their immediate vicinity or over large distances. Cytokines are produced by a wide range of cells, including immune cells and endothelial cells. They include interferons, interleukins, and tumour necrosis factors.

A bee swarm

Stallion exposing receptors

Pheromones are chemical signals released into the external environment. They are widely used by many animals, especially social insects and mammals. Pheromones have different purposes (e.g. aggression, aggregation, reproduction, territoriality) but all act to generate a specific response in members of the same species (conspecifics). A gland at the base on the honeybee abdomen secretes Queen Mandibular Pheromone, which attracts worker bees to swarm. This behaviour is important when establishing a new hive. In mammals, pheromones are used to signal sexual receptivity and attract mates. Special receptors in the nasal cavity detect the pheromones. Mammals often curl the upper lip (flehmen) to expose the receptors.

1. Compare how hormones are transported in plants and animals:

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4. How do pheromones differ from the other types of chemical signals?

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2. (a) Identify a type of long lasting chemical signal: (b) Identify a type of short lasting chemical signal: 3. What are the benefits of using both short and long acting signals when coordinating physiology and behaviour?

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68 What is Signal Transduction?

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98

Key Idea: Signal transduction is the conversion of an external signal to a functional change within the cell through a series of biochemical reactions. Signal transduction is the process by which molecular signals are transmitted from outside the cell to inside, bringing about a cellular response. The transduction involves an external signal molecule binding to a receptor and triggering a series

of biochemical reactions, which lead to a specific cellular response. The series of biochemical reactions is often called a cascade and usually involves phosphorylation (charging) of a number of molecules in a sequence. The type of response varies and may include a change in metabolism (activating a pathway), gene expression (to produce a specific protein), or membrane permeability (to allow entry of specific molecules).

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An overview of signal transduction

Plasma membrane

Extracellular fluid

Cytoplasm

1. Reception

2. Transduction

3. Response Activation of cellular response within the target cell

Receptor

Many molecules are involved in a signal transduction pathway

Signal molecule

Signal transduction can be broken into three main steps:

ffReception: An extracellular signal molecule binds to its receptor on a target cell.

ffTransduction: The activated receptor triggers a chain of biochemical events within the cell. Many different enzymes are involved, and the entire reaction is often called a signalling cascade.

ffResponse: The signal cascade results in a specific cellular response.

Gibberellic acid activation of a-amylase: An example of a cellular response

In plants, the hormone gibberellic acid (GA) is involved in seed germination. GA acts as a signal molecule to stimulate the production of the enzyme a-amylase. The a-amylase hydrolyses (breaks down) starch into simple sugars, which the plant can use. GA

Myb

GA receptor

GA binds to a receptor on the plasma membrane.

GA receptor signals for the production of a transcription factor called Myb protein.

Myb is needed to produce α-amylase

Myb protein binds to DNA and activates transcription of a-amylase enzyme.

α-amylase

a-amylase is produced and hydrolyses the starch in the seeds into simple sugars.

(b)

(c)

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(a)

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1. Name the three stages of signal transduction and describe what occurs at each stage:

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69 Types of Signal Transduction

Key Idea: The majority of cell signals bind to extracellular receptors to exert their effect. However some cell signals are able to pass through the plasma membrane and bind directly to intracellular receptors within the cell to exert their effect. Cell receptors fall into two broad classes. Extracellular receptors bind signal molecules outside of the cell. The

Hydrophobic signal molecules are received by intracellular receptors

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Hydrophilic signal molecules are received by extracellular receptors

signal molecule does not have to pass across the plasma membrane to cause a cellular response. Most cell receptors are extracellular receptors. Intracellular receptors bind signal molecules that have passed into the cell directly across the plasma membrane. Intracellular receptors may be located in the cytoplasm or on the nucleus.

Cortisol is involved in glucose metabolism and response to stress

Adrenaline accelerates heart rate and is involved in the fight or flight response

Extracellular fluid

The first messenger (signal molecule) binds to the receptor protein

Extracellular fluid

Signal molecule

Enzyme

Receptor protein

Plasma membrane

Plasma membrane

Protein subunit from the receptor protein activates the enzyme

Second messenger

Lipid soluble signal molecule passes across the plasma membrane

Signal binds to receptor to form a receptor/signal complex

Active enzyme produces a second messenger

Receptor/signal complex acts as a transcription factor, binding to DNA to begin transciption

The second messenger triggers a cascade of phosphorylation events leading to a cellular response

DNA

Nucleus

P

The protein product produced alters the cell’s activity

Cell response

Cell response Cytoplasm

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Hydrophobic signal molecules diffuse freely across the plasma membrane and into the cytoplasm of target cells. In the example above the signal molecule binds to a receptor in the cytoplasm to form a receptor/signal complex. The complex moves to the cell nucleus where it acts as a transcription factor to alter gene transcription, thus controlling the expression of specific genes. Steroid hormones, such as cortisol and sex hormones, are examples of hydrophobic signal molecules.

CL

Hydrophilic signal molecules are water soluble and cannot cross the plasma membrane. They cannot exert their effect without the aid of an extracellular receptor. Hydrophilic signals include water soluble hormones such as adrenaline. The signal molecule is called the first messenger. When it binds, the extracellular receptor is activated, triggering a sequence of biochemical reactions, including activation of a second messenger. As a consequence of these sequential reactions, the original signal is amplified, bringing about a cellular response.

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Cytoplasm

WEB

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100 1. Describe the differences between an intracellular receptor and an extracellular receptor:

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2. What must a signal molecule do in order to activate a receptor?

3. In terms of their ability to cross the plasma membrane, describe the difference between a hydrophobic signal molecule and a hydrophilic signal molecule:

4. (a) Outline the process when signal transduction occurs via an extracellular receptor:

(b) Describe the differences between a first messenger and a second messenger:

5. Outline the process when signal transduction occurs via an intracellular receptor:

6. The diagram on the right represents a cell signalling process.

(a) Does this diagram represent an extracellular or intracellular signalling process? Explain your answer:

Plasma membrane

Cell response

(b) What type of receptor is B? (c) What does A represent? (d) Would A be hydrophobic or hydrophilic? Explain your answer:

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B

CL

A

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70 Apoptosis: Programmed Cell Death

Key Idea: Apoptosis is a controlled cell suicide that maintains cell numbers and sculpts body parts during development. It is a tightly regulated process. Apoptosis, also called programmed cell death (PCD), is a natural and necessary mechanism in multicellular organisms to trigger the death of a cell. Apoptosis helps to maintain adult cell numbers and is a defence against damaged or dangerous

cells, such as virus-infected cells and cells with DNA damage. Apoptosis also has a role in “sculpting” embryonic tissue during development, e.g. in the formation of digits in developing embryos and resorption of the larval tail during amphibian metamorphosis. Apoptosis is tightly controlled and can be initiated through intrinsic (mitochondrial) or extrinsic (death receptor) pathways in response to signal molecules.

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An overview of apoptosis ffApoptosis is a controlled process of cell suicide. It occurs in response to

specific cell signals and involves an orderly series of biochemical events.

ffThe cell and its nucleus shrink and there is an orderly dissection of chromatin by endonucleases.

ffDeath is finalised by a rapid engulfment of the dying cell by phagocytosis. This safely disposes of the remains of the cell.

Ed Uthman

The cell shrinks and forms blebs. The nucleus degrades and the chromatin forms clumps.

Blebs

Cells that are stressed, damaged, or triggered by signal molecules begin apoptosis.

Nucleus

Organelle

The nucleus collapses but some organelles are not affected. The cell sends signals to attract macrophages.

The nucleus breaks up into spheres, the DNA breaks up, and the cell fragments.

The cell breaks into numerous apoptotic bodies. Macrophages remove them by phagocytosis.

Apoptotic body

In humans, the mesoderm tissue initially formed between the fingers and toes is removed by apoptosis. Forty one days after fertilisation (top), the digits of the hands and feet are webbed, appearing paddle-like. Apoptosis selectively destroys this superfluous webbing, sculpting them into digits, which can be seen in later stages of development (above).

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2. What is the role of apoptosis in the normal functioning of the body?

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1. What is apoptosis?

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Activating apoptosis

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ffThe initiation of apoptosis is very tightly regulated because once it has begun it cannot be stopped. Two commonly studied activation pathways are the intrinsic or mitochondrial pathway and the extrinsic or death receptor pathway.

ffThe intrinsic pathway is dependent on factors released from mitochondria. It is activated from within the cell in response to cellular stress. Triggers include deprivation of growth factors, DNA damage, and low oxygen levels.

ffThe extrinsic pathway is activated when extracellular signals bind to transmembrane death receptors on the cell's surface.

An important extracellular signal molecule for this pathway is Tumour Necrosis factor (TNF) a cytokine produced by activated macrophages. Most human cells have TNF receptors.

ffBoth these apoptotic pathways depend on caspases (protein digesting enzymes). Caspases cleave specific proteins in the

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cytoplasm or nucleus to bring about apoptosis. The extrinsic pathway can also trigger the intrinsic pathway by activating BID protein, which is involved in the regulation of apoptosis. BID causes changes in the mitochondrion that lead to caspase activation. An overview of both pathways is shown below. Note that each pathway activates distinct initiator caspases, but both use the same effector caspases.

Extrinsic pathway

Intrinsic pathway

Extracellular fluid

Signal molecule, e.g. TNF Death receptor

Plasma membrane Cytoplasm

Inactive caspase 8

• Lack of oxygen • Insufficient growth factors • Irreversible DNA damage

BID

Active caspase 8

Mitochondrion

Initiation caspase 9

Active caspase 9

Effector caspases 3, 6, 7

Apoptosis

3. Why must apoptosis be tightly regulated?

4. (a) Name the two activation pathways for apoptosis:

(b) Briefly outline the differences between the two pathways:

(c) What do the two pathways have in common?

5. Describe the role of caspases and BID in apoptosis:

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N AS OT SR F OO OR M US E

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71 Malfunctions of Apoptosis

Key Idea: An imbalance in the factors controlling apoptosis can alter the normal apoptosis rate and cause disease. Apoptosis serves a critical role in removing damaged or abnormal cells before they can proliferate. It is not surprising then that malfunctions in this process can lead to disease. Cancers typically involve a reduction in the normal rate of apoptosis so that damaged cells continue to proliferate instead

of undergoing a programmed cell death. In contrast, some neurodegenerative and immune system diseases, including Alzheimer's and autoimmune diseases, are associated with abnormally high rates of apoptosis. This process may be triggered by infection. For example, HIV-induced apoptosis of the immune system's T-helper cells leads to immune system failure and an inability to control viral replication.

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Reduction in rates of apoptosis can cause cancer Normal cell division

Unrepaired cell damage

Cell division in cancer

Cancerous cells can disrupt and evade normal apoptotic pathways, continuing to divide and forming tumours.

Apoptosis

Tumour suppressor genes, e.g. p53, normally halt cell division of DNA damaged cells until the damage is repaired. If the damage cannot be repaired, apoptosis is triggered.

Herpesvirus

First mutation

Second mutation

Third mutation

No apoptsis and uncontrolled growth

Cancerous cells may inhibit the expression of p53 or increase the expression of signals to halt apoptosis. As many as 50% of all human tumours contain p53 mutations.

These brain sections, normal (left) vs Alzheimer's (right) show how much nervous tissue has been lost.

Some viruses inhibit apoptosis Some viruses, such as the Herpesvirus, inhibit apoptosis by inhibiting caspase activity within their host cell. This prevents premature death of host cells and allows the virus to persist in the body.

Fourth or later mutation

Increased apoptosis can causes disease People with Alzheimer's disease have high levels of the enzyme GSK3. GSK3 promotes the intrinsic apoptosis signalling pathway and contributes to the abnormally high rates of nerve cell apoptosis seen in the brains of Alzheimer's patients (above).

HIV viruses budding from a T cell

Some viruses stimulate apoptosis HIV is a virus that stimulates apoptosis to its benefit. HIV accelerates the apoptotic destruction of T-helper cells through several mechanisms, include direct toxicity and signalling by viral proteins.

Self sufficient in

growth signals 1. Explain how a decreased rate of apoptosis can lead to cancer: Insensitive to anti-growth signals

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Evades apoptosis

2. A number of Herpesviruses are recognised oncogenic (cancer-forming) agents. How might they exert this effect? Unlimited potential for cell division Sustained formation of new lood vessels Tissue invasion and metastasis

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3. Explain how increased rates of apoptosis can lead to loss of tissue mass and system failure:

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72 KEY TERMS AND IDEAS: Did You Get It?

1. Match each term to its definition, as identified by its preceding letter code. apoptosis

A A molecule that relays signals from receptors on the cell surface to target

extracellular receptor

molecules inside the cell.

B An extracellular substance that binds to an extracellular receptor and initiates

extrinsic pathway

intracellular activity.

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C In apoptosis, this pathway is initiated by a signal binding to a death receptor.

first messenger

hydrophilic signal

D A receptor located within a cell.

E A chemical molecule that binds to a cellular receptor and brings about a change within the cell.

hydrophobic signal

intracellular receptor intrinsic pathway

F The process by which an external cell signal binds to a receptor and the receptor

triggers a series of biochemical reactions in the cell to cause a specific response.

G A signal molecule that is able pass directly through the plasma membrane.

H The process of programmed cell death in multicellular organisms in which the cell goes through a series of orderly chemical events leading to death.

second messenger signal molecule

signal transduction

I In apoptosis, this pathway is initiated from within a cell in response to cellular stress, such as low oxygen.

J A cell that responds in a particular way to a specific signal molecule.

K A signal molecule which cannot pass directly through the plasma membrane and must bind to an extracellular receptor.

target cell

L A receptor that spans the plasma membrane.

2. (a) The molecules labelled A-C are signalling molecules. Identify the signal molecule that will bind to the receptor shown:

A

B

C

(b) What prevents the other two signal molecules from binding to this receptor?

Receptor

(c) Why is it important that not all cells react to every signal molecule?

(a) What is this process called?

(b) Did this process occur normally in the development of the person shown right?

(c) What is the purpose of this process in adults?

(d) Describe a consequence of malfunctions in this process in adults:

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3. Webbing of the toes and fingers in early human fetal development is normal. At 16 weeks a process occurs to remove the webbing.

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Responding to antigens

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Unit 3 Outcome 2

Key terms

Antigens and pathogens

adaptive (=specific) immune response

Key knowledge

Activity number

antibody (=immunoglobulin)

c

1

Explain what is meant by an antigen. Distinguish between non-self antigens, self-antigens, and allergens and explain the importance of the distinction.

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antigen

c

2

Explain how the body distinguishes self from non-self. Include reference to the role of the major histocompatibility complex (MHC) and its role in self-recognition.

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cellular (cell mediated) immunity

c

3

Distinguish between cellular and non-cellular pathogens. Describe the role of pathogens as a source of non-self antigens.

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clonal selection

c

4

Describe the range of physical and chemical defences against pathogen invasion in plants and animals. Include reference to physical barriers, chemical defences, and microbiological barriers (the body's normal microflora).

B cell (=B lymphocyte)

dendritic cell histamine

75 76

humoral immunity immunity

immunological memory infection

inflammation

innate (=non-specific) immune response

National Cancer Institute

interferon

The innate immune response

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lymphocyte

macrophage mast cell MHC

phagocyte

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Outline the general structure of the lymphatic system, distinguishing between primary lymphoid organs and secondary lymphoid tissues.

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Explain the role of the lymphatic system in the immune response. Include reference to the role of lymph nodes as the site of antigen recognition by lymphocytes and the importance of the lymphatic system in transporting immune cells (e.g. dendritic cells) around the body.

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T helper cell thymus

The adaptive immune response Key knowledge

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T cytotoxic cell

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Describe the components of the adaptive (specific) immune response (cellular and humoral immunity), including the role of specificity and memory.

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Recognise clonal selection and the basis of immunological memory. Explain how the immune system is able to respond to the large range of potential antigens.

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Describe the actions of B-lymphocytes (B cells) and their antibodies in humoral immunity. Include reference to antibody structure and how this relates to function.

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Describe the actions of T-lymphocytes in cellular immunity to include T helper and T cytotoxic (killer) cells.

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T cell (=T lymphocyte)

Describe the features and roles of the components of the innate (or non-specific) immune response to antigens, including the steps in inflammatory response. Include reference to the role of: (a) Complement proteins in the plasma in stimulating phagocyte activity. (b) Mast cells and the release of histamine. (c) Macrophages and neutrophils in phagocytosis. (d) Dendritic cells in processing and presenting antigenic material.

The role of the lymphatic system

primary lymphoid organs

secondary lymphoid tissue

Activity number

Key knowledge

lymph

lymph node

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73 The Nature of Antigens

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and tissues. Antigens that originate from within the body are called self-antigens. Normally, because of the development of self-tolerance, the body recognises and does not attack its own tissues. However, in some instances, the immune system may mistakenly destroy its own tissues. Such a response is called an autoimmune disorder. Allergens are a specific type of antigen, they produce a vigorous hypersensitive allergic response.

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Key Idea: Antigens are substances capable of producing an immune response. It is important that the body can distinguish its own tissues from foreign material so that it does not attack itself. An antigen is any substance that evokes an immune response in an organism. Most antigens are non-self antigens, i.e. they are foreign and originate from outside the organism. Sometimes an organism will react to its own cells

Distinguishing self from non-self ffEvery type of cell has unique protein markers (antigens)

on its surface. The type of antigen varies greatly between cells and between species. The immune system uses these markers to identify its own cells (self) from foreign cells (non-self). If the immune system recognises the antigen markers, it will not attack the cell. If the antigen markers are unknown, the cell is attacked and destroyed.

ffIn humans, the system responsible for this property is

the major histocompatibility complex (MHC). The MHC is a cluster of tightly linked genes on chromosome 6. These genes code for protein molecules (MHC antigens) that are attached to the surface of body cells. The main role of MHC antigens is to bind to antigenic fragments and display them on the cell surface so that they can be recognised by the cells of the immune system.

ffClass I MHC antigens are found on the surfaces of

almost all human cells. Class II MHC antigens occur only on macrophages and B-cells of the immune system.

Genes for producing the MHC antigens Class I MHC

Class II MHC

MHC surface proteins (antigens) provide a chemical signature that allows the immune system to recognise the body’s own cells

Tolerance towards foreign bodies ffThe human body has a very large population of resident microbes. Under normal conditions, E.coli in the gut form a protective layer preventing the colonisation of pathogenic bacteria. The microbial cells have foreign antigens but they are not attacked by the immune system because tolerance (the prevention of an immune response) has developed.

E.coli

ffDuring pregnancy, specific features of the self recognition system are suppressed to allow the mother to tolerate a nine month relationship with a foreign body (the fetus).

Intolerance to tissue transplants

The MHC is responsible for the rejection of tissue grafts and organ transplants. Foreign MHC molecules on the transplanted tissue are viewed as antigenic, causing the immune system to respond and the tissue to be rejected. To minimise rejection, attempts are made to match the MHC of the organ donor to that of the recipient as closely as possible. Immunosuppressant drugs are also used to minimise the immune response.

Chromosome 6

Kidney transplant

1. (a) What is an antigen?

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(c) Why is it important that the body detects foreign antigens?

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(b) Distinguish between non-self antigens, self antigens, and allergens:

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Types of antigens Self antigens

Any foreign material provoking an immune response is termed a non-self antigen. Disease-causing organisms (pathogens) such as bacteria, viruses, and fungi are nonself antigens. The body recognises them as foreign and will attack and destroy them before they cause harm.

The body is usually tolerant of its own antigens. However, sometimes the self-tolerance system fails and the body attacks its own cells and tissues as though they were foreign. This can result in an autoimmune disorder in which tissue is destroyed, grows abnormally, or changes in function.

Allergens Antigens that cause allergic reactions are called allergens. An allergic reaction is a very specific type of immune response in which the immune system overreacts to a normally harmless substance. An allergic response can produce minor symptoms (itching, sneezing, rashes, swelling) or life-threatening anaphylaxis (respiratory and cardiovascular distress).

CDC

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Non-self antigens

Autoimmune disorders, such as multiple sclerosis and rheumatoid arthritis, may be triggered by infection by a pathogen. Similarity in pathogen and self antigens is thought to be behind this failure of self recognition.

Common allergens include dust, chemicals, mould, pet hair, food proteins, or pollen grains.

Pathogens have ways of avoiding detection. Mutations result in new surface antigens, delaying the immune response and allowing the pathogen to reproduce in its host undetected for a time (e.g. the flu virus, above). Some pathogens, e.g. the malaria-causing Plasmodium, switches off its surface antigens in order to enter cells undetected.

Type 1 diabetes is the result of autoimmune destruction of the insulinproducing pancreatic cells. Patients must inject insulin to maintain normal blood glucose levels.

Kent Pryor

Influenzavirus

The swelling on the foot in the left of the photograph is a result of an allergic reaction to a bee sting.

2. How can pathogens avoid detection by the immune system?

3. (a) What is the nature and purpose of the major histocompatibility complex (MHC)?

(b) Why is a self-recognition system important?

4. (a) What is immune tolerance?

(b) When might tolerance to foreign antigens be beneficial or necessary?

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(b) Normally non-antigenic substances:

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5. Using examples, describe what happens when the body develops an inappropriate response to: (a) Self-antigens:


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74 Cellular and Non-Cellular Pathogens

Key Idea: Pathogens are infectious agents that spread between hosts and cause disease. Pathogens can be cellular (living) or non-cellular (non-living). Pathogens are disease-causing agents. They, or their toxins, act as antigens and provoke an immune response from the host. It is the activity of the immune system that causes the symptoms we associate with disease, such as fever and inflammation. Cellular pathogens are living organisms. They

have all the cellular machinery to carry out their life process and reproduce. Examples include bacteria and eukaryotic pathogens (fungi, algae, protozoa, and parasitic worms). Noncellular pathogens are regarded as non-living. They lack the cellular machinery required to carry out metabolism on their own and make use of the host's metabolism to reproduce. Viruses and prions (infectious agents composed entirely of protein) are non-cellular pathogens.

Mycobacterium tuberculosis (MTB)

Ute Frevert, Plos

CDC: Janice Haney Carr

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Cellular pathogens

Plasmodium sporozoite

Bacterial pathogens

Fungal pathogens

Protistan pathogens

Pathogenic bacteria are responsible for some of the world's most devastating plant and animal diseases. Bacterial disease can be transmitted through food, water, air, or by direct contact. Pulmonary tuberculosis (TB) is caused by Mycobacterium tuberculosis (above). The pathogen is spread through the air when infectious people cough, sneeze, talk, or spit. Other notable bacterial diseases include cholera, food poisoning, tetanus, and whooping cough.

Pathogenic fungi are more common in plants than in animals. They spread by spores and the infections they cause are generally chronic (long-lasting) infections because fungi grow relatively slowly. Fungal infections often affect the skin, hair, nails, genital organs, and inside of mouth. Infection occurs after coming into contact with spores. Many fungal diseases are not serious (e.g. infected toenails, above) but others (e.g. those infecting the blood) can be fatal.

Protistans are a large and diverse group of eukaryotes. A number of species are significant pathogens of animals or plants. Pathogenic protists have very complex life cycles, often involving a number of different hosts and several different life stages. For example, malaria is caused by protozoan parasites called plasmodia. Plasmodia have a life cycle involving two hosts, Anopheles mosquitoes (the vector) and humans. Humans become infected when bitten by infected mosquitoes.

Non-cellular pathogens

A virus is a highly infectious pathogen that infects living cells (including bacterial cells) and uses the cell's metabolic machinery to replicate. Viruses are responsible for a number of human diseases including the common cold, flu, HIV/AIDS, polio, hepatitis, and childhood diseases such as measles, chickenpox, and rubella. The severity of the disease and mode of transmission varies greatly between different viruses.

Prions are misfolded infectious proteins that have pathogenic properties. They contain no genetic material so do not replicate in the usual way. Instead, an infectious prion binds to a normal protein and causes it to change shape and become infectious. Prions cause degenerative nervous diseases in mammals including scrapie in sheep, BSE in cattle, and kuru in humans.

Rabies virus

Ebolavirus

Normal Normalprotein protein

Infectious protein Infectious protein

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1. Explain how pathogens cause disease:

3. List four ways in which pathogens can be transmitted between hosts:

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2. Distinguish between cellular and non-cellular pathogens and give examples of each:

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75 The Body's Defences: An Overview

Key Idea: The human body has a tiered system of defences that provides resistance against disease. The body has a suite of physical, chemical, and biological defences against pathogens, collectively called resistance. The first line of defence consists of external barriers to prevent pathogen entry. If this fails, a second line of defence targets any foreign bodies that enter. Lastly, the specific immune response provides targeted defence against the pathogen. The defence responses of the body fall into two

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broad categories, the innate and the adaptive immune responses. The innate (non-specific) response (the first and second lines of defence) protects against a broad range of non-specific pathogens. It involves blood proteins (e.g. complement) and phagocytic white blood cells. The adaptive (or specific) immune response (the third line of defence) is specific to identified pathogens. It involves defence by specific T cells (cellular immunity) as well as antibodies, which neutralise foreign antigens (humoral immunity).

Most microorganisms find it difficult to get inside the body. If they succeed, they face a range of other defences.

The natural populations of harmless microbes living on the skin and mucous membranes inhibit the growth of most pathogenic microbes

Microorganisms are trapped in sticky mucus and expelled by cilia (tiny hairs that move in a wavelike fashion).

Intact skin

1st line of defence

The skin provides a physical barrier to the entry of pathogens. Healthy skin is rarely penetrated by microorganisms. Its low pH is unfavourable to the growth of many bacteria and its chemical secretions (e.g. sebum, antimicrobial peptides) inhibit growth of bacteria and fungi. Tears, mucus, and saliva also help to wash bacteria away.

Mucous membranes and their secretions:

2nd line of defence

Antimicrobial substances

A range of defence mechanisms operate inside the body to inhibit or destroy pathogens. These responses react to the presence of any pathogen, regardless of which species it is. White blood cells are involved in most of these responses. It includes the complement system whereby plasma proteins work together to bind pathogens and induce inflammation to help fight infection.

3rd line of defence

Inflammation and fever 40°C

Phagocytic white blood cells

37°C

Eosinophils: Produce toxic proteins against certain parasites, some phagocytosis

Basophils: Neutrophils, macrophages: Release heparin (an These cells engulf and anticoagulant) and histamine destroy foreign material which promotes inflammation (e.g. bacteria)

Antibody

Specialised lymphocytes

T-cells: Recognise specific antigens and activate specific defensive cells.

B-cells: Recognise specific antigens and divide to form antibody-producing clones.

1. Distinguish between the adaptive and innate immune responses:

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Harmless microbes colonise certain areas of the body (e.g. skin, gut, mouth, nose and throat) and prevent pathogens establishing.

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Antimicrobial chemicals are present in many bodily secretions, including tears. Tears also wash away contaminants from the eyes.

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Intact skin (above) provides a physical barrier to stop pathogens entering the body. Cuts or abrasions allow pathogens to enter.

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Helena Paffen CC3.0

Once the pathogen has been identified by the immune system, lymphocytes launch a range of specific responses to the pathogen, including the production of defensive proteins called antibodies. Each type of antibody is produced by a B-cell clone and is specific against a particular antigen.

Lining of the respiratory, urinary, reproductive and gastrointestinal tracts

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Key Idea: Plants and animals have chemical defences to defend against pathogens. Some mechanisms are always present while others are stimulated by an attack. Living organisms are under constant attack from pathogens. As a result plants and animals have evolved a wide range of chemical defences to protect themselves from pathogens and

limit the damage they can do. Some defence mechanisms are always present, while others are stimulated only by an attack. The chemical defences of plants not only protect them from attack by pathogens, but also stop animals eating them or inhibit the growth of other plants.

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Chemical defences in animals

Vertebrate defences

Invertebrate defences

Antimicrobial substances: Chemicals (e.g. lactoferrin) are secreted from the skin and other body fluids, and also by some white blood cells. These kill pathogens or inhibit their growth.

Antimicrobial peptides: Defensins are abundant in phagocytes and small intestinal mucosa of mammals. They disrupt the plasma membrane of bacterial cells causing death.

Lysozyme is an enzyme found in tears, saliva, milk, and mucus It damages bacterial cell walls and the bacteria dies.

Image: Ccroberts

Inflammatory response: Release of the chemicals heparin and histamine (above) promote inflammation to limit pathogen spread.

ProPO system: The presence of microbial compounds initiates the prophenoloxidase (proPO) defence system. proPO produces a cascade, and the final product, melanin, encases and kills the pathogen. The system is present in most invertebrates.

Lytic system: Enzymes such as lysozyme breakdown (by hydrolysis) bacterial cell walls, killing the bacterial cells.

PHOTO: Kristof A. & Klussmann-Kolb A. cc 2.0

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76 Chemical Defences In Plants and Animals

Aeolidiella stephanieae, a sea slug

Antimicrobial peptides: The antimicrobial peptide defensin binds to the cell membrane of pathogens, and causes damage by puncturing the membrane. The pathogen becomes "leaky" and dies.

1. (a) Describe the advantage of having multiple (non-specific) defence responses:

(b) Describe a disadvantage of having only general (non-specific) defence responses:

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2. Compare and contrast the non-specific defences of vertebrate and invertebrate animals:

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Chemical defences in plants

Passive defences

Once infected, a plant responds actively to prevent any further damage. Active defences are invoked only after a pathogen has been recognised, or after wounding or attack by a herbivore. This makes biological sense because active defences are costly to produce and maintain. Active defences work through a variety of mechanisms including slowing pathogen growth, puncturing the cell wall, disrupting metabolism, or killing cells by release of reactive oxygen species such as hydrogen peroxide (H2O2).

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Passive defences are always present and are not the result of contact with a pathogen or grazer. Plants have both physical and chemical defences to deter pathogens. For example, the thick waxy surface of many leaves (right) acts as a physical barrier to limit pathogen entry. However, if the physical defence is breached, the chemical defences protect the plant against further damage.

Active defences

Invading fungal hypha

hypha

cell

cell wall

Many plants repair their cell wall when invaded by pathogens

The powdery mildew infecting this plant is a fungus

EII

Many plants produce an enzyme-activated hypersensitive response when invaded by pathogens. This leads to the production of reactive nitric oxide and cell death. Cell death in the infected region limits the spread of the pathogen.

Many plants produce a range of antimicrobial and antifungal chemicals and enzymes to kill or inhibit the growth of pathogens. Some of these compounds cover the surface of the plant, killing pathogens before they enter the plant. Other compounds act internally. Many herbs have antimicrobial properties. These compounds are sometimes extracted for human use.

Sealing off infected areas gives rise to abnormal swellings called galls (oak gall, left and bulls-eye galls on a maple leaf, right). These galls limit the spread of the parasite or the infection in the plant.

3. (a) Distinguish between passive and active defence mechanisms in plants:

(b) Why are most plant defensive chemicals produced only after a pathogen is detected?

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4. How are galls effective in reducing the spread of infection in some plants?

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5. What similarities are there between the active defence mechanisms of plants and the immune responses of animals?


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77 The Innate Immune Response

Key Idea: The innate immune response provides a rapid response to contain and destroy pathogens. Inflammation is an important part of the response. The innate immune system provides protection against a pathogen, even if it has never encountered it before. The innate response is very fast and provides general protection (it is not antigen specific), but does not provide long lasting

immunity. Many different cells and processes are involved. The primary outcome is to destroy and remove the cause of infection. This is achieved through containing the infection through inflammation and then recruitment of immune cells to destroy the pathogen. During this process a series of biochemical reactions (the complement system) are activated to destroy the pathogen and recruit immune cells to the site.

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Phagocytic cells of the innate immune system

Volker Brinkmann PLOS cc 2.5

Obli cc 2.0

A macrophage stretches its 'arms' to engulf pathogens

A neutrophil engulfs bacterial cells

Artist's rendering of activated dendritic cell

Macrophage

Neutrophil

Dendritic cell

Macrophages are very large and are highly efficient phagocytes. They are found throughout the body and move using an amoeboid movement (above) to hunt down and destroy pathogens. Macrophages also have a role in recruiting other immune cells to an infection site.

Neutrophils are the most abundant type of phagocyte and are usually the first cells to arrive at the site of an infection. They contain toxic substances that kill or inhibit the growth of bacteria and fungal pathogens. Neutrophils release cytokines which amplify the immune response and recruit other cells to the infection site.

Dendritic cells are present in tissue that are in contact with the external environment (e.g. skin, and linings of the nose, lungs, and digestive tract). They act as messengers between the innate and adaptive immune system by presenting antigen materials to the T cells of the immune system.

Other cells and processes of the innate immune response

Complement component 3 (C3)

Mast cells

Complement proteins

The process of inflammation

Mast cells contain a lot of histamine, a chemical involved in both inflammation and allergic responses. When activated, histamine is released from the mast cell causing the blood vessels to dilate and become leaky. The increased permeability allows phagocytes to reach the site of infection.

The complement system comprises a number of different proteins. The proteins circulate as inactive precursors until they are activated. Complement proteins have three main roles: phagocytosis, attracting macrophages and neutrophils to the infection site, and rupturing the membranes of foreign cells.

The inflammatory process is a protective response to pathogen invasion. It has several functions: (1) to destroy the cause of the infection and remove it and its products from the body; (2) if this fails, to limit the effects on the body by confining the infection to a small area; (3) replacing or repairing tissue damaged by the infection.

(a) Macrophages:

(b) Neutrophils:

(c) Dendritic cells:

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1. Outline the role of the following phagocytes in the innate immune response:

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The inflammatory response Bacteria entering on knife or other sharp object.

Blood clot forms

Chemicals (e.g. histamines and prostaglandins) are released by damaged cells, attracting more and more phagocytes to the infection.

Bacterium

Red blood cells

Capillary wall

Phagocytes stick to capillary walls

Neutrophil

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Epidermis Dermis

Bacteria

Macrophage

Subcutaneous tissue

Blood vessels increase diameter (vasodilation) and permeability.

An abscess starts to form after a few days. This collection of dead phagocytes, damaged tissue and various body fluids is called pus.

Bacteria are engulfed and destroyed by phagocytes (macrophages and neutrophils).

Phagocytes squeeze between cells making up blood vessel walls.

Stages in inflammation

Increased diameter and permeability of blood vessels

Blood vessels increase their diameter and permeability in the area of damage. This increases blood flow to the area and allows defensive substances to leak into tissue spaces.

Phagocyte migration and phagocytosis

Within one hour of injury, phagocytes appear on the scene. They squeeze between cells of blood vessel walls to reach the damaged area where they destroy invading microbes.

Tissue repair

Functioning cells or supporting connective cells create new tissue to replace dead or damaged cells. Some tissue regenerates easily (skin) while others do not at all (cardiac muscle).

2. What role does the complement system play in immunity?

3. Outline the three stages of inflammation and identify the beneficial role of each stage: (a)

(b)

5. Why does pus form at the site of infection?

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4. What role do mast cells play in inflammation?

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(c)


78 Phagocytes and Phagocytosis

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Key Idea: Phagocytes are mobile white blood cells that ingest microbes and digest them by phagocytosis. All types of phagocytes (e.g. neutrophils, dendritic cells, and macrophages) are white blood cells. These specialised cells have receptors on their surfaces that can detect antigenic material, such as microbes. They then ingest the microbes

Opsonins

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Detection and interaction Microbe coated in opsonins is detected by the phagocyte and attaches to it. Opsonins are molecules in the blood that coat foreign material (e.g. a bacterial cell), marking it as a target for phagocytosis.

and digest them by phagocytosis. As well as destroying microbes, phagocytes also release cytokines that help to coordinate the overall response to an infection. Macrophages and dendritic cells also play a role in antigen presentation in processing and presenting antigens from ingested microbes to other cells of the immune system (opposite).

Engulfment The opsonin markers trigger engulfment of the microbe by the phagocyte. The microbe is taken in by endocytosis.

Microbe

Opsonin receptors

Phagosome forms A phagosome forms, enclosing the microbe in a membrane.

Phagosome

Lysosome

Fusion with lysosome Phagosome fuses with a lysosome containing powerful antimicrobial proteins. The fusion forms a phagolysosome.

Nucleus

Digestion The microbe is broken down into its chemical constituents.

Phagolysosome

Discharge Indigestible material is discharged from the phagocyte.

The interaction of microbes and phagocytes

Some microbes kill phagocytes.

Dormant microbes may hide inside phagocytes.

Microbes enter phagocytes and evade the immune response.

Some microbes kill phagocytes

Microbes evade immune system

Dormant microbes hide inside

Some microbes produce toxins that can actually kill phagocytes, e.g. toxin-producing staphylococci and the dental plaque-forming bacteria Actinobacillus.

Some microbes can evade the immune system by entering phagocytes. The microbes prevent fusion of the lysosome with the phagosome and multiply inside the phagocyte, almost filling it. Examples include Chlamydia, Mycobacterium tuberculosis, Shigella, and malarial parasites.

Some microbes can remain dormant inside the phagocyte for months or years at a time. Examples include the microbes that cause brucellosis and tularemia.

1. Identify the white blood cells capable of phagocytosis:

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2. Explain the role of opsonins and phagocyte receptors in enhancing phagocytosis:

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3. Explain how some microbes can overcome phagocytic cells and use them to their advantage:

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79 Processing Antigens

Key Idea: Antigen processing prepares and displays antigens for presentation to the T-cells of the immune system. Antigen presenting cells (APCs) process and present antigens for recognition by T-cells. During antigen processing, the APC digests the foreign antigen into smaller peptide fragments. These fragments are then displayed on the surface of the

APC by MHC receptors. The immune response evoked by the T-cells depends on which MHC receptor (MHCI or MHCII) is activated. Antigen presentation is necessary for T-cells to recognise infection or abnormal growth and activate other cells of the immune system. Dendritic cells, macrophages, and B-cells are APCs. Antigen binding site

The role of MHC receptors

Antigen binding site

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Recall there are two types of MHC receptors, class I and class II (right). Both have similar functions in that they display antigenic peptides on cell surfaces so antigens can be recognised and processed by the T-cells of the immune system. T-cells can only recognise antigenic peptides if they are displayed by the MHC receptors. MHC receptors presenting no foreign antigenic peptides are ignored by T-cells, because they are signalling that the cell is healthy. Only MHC receptors with foreign antigenic peptides bound to them will attract T-cells and evoke an immune response. The source of the antigenic peptides bound to each class of MHC receptor differs. Class I MHC receptors display antigenic peptides of intracellular parasites such as viruses. Class II MHC receptors display antigenic peptides originating from outside of the cell (such as those from ingested microbes).

Class I MHC Intracellular antigens, e.g. viral proteins

Class II MHC Extracellular antigens, e.g. proteins from phagocytosed microbes

An overview of antigen processing

The diagram on the right represents antigen processing of an extracellular peptide antigen via a class II MHC receptor. An APC encounters an antigen.

Class II MHC receptor

T helper cell

The antigen is engulfed via phagocytosis and digested into short peptide fragments.

Class II MHC receptors bind the fragments and form a MHC-antigen complex.

The MHC-antigen complex is displayed on the surface of the APC.

Class II MHC receptor/antigen complex

Antigen

Antigen peptide fragment

A receptor on the T helper cell recognises the peptide as foreign. It binds and a series of events stimulate the adaptive immune response.

Class II MHC

APC (dendritic cell)

1. What is the purpose of antigen processing?

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3. Describe the differences between class I and class II MHC receptors:

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2. Why do MHC receptors with no antigenic peptide bound not cause an immune response?

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80 The Lymphatic System

Key Idea: Defensive white blood cells are transported in lymph through the lymphatic system and are concentrated in the lymph nodes. The lymphatic system is a network of tissues and organs that collects the tissue fluid leaked from the blood vessels and returns it to the heart. The lymphatic system has an important

role in immunity because the fluid transported around the body by the lymphatic system (lymph) is rich in infectionfighting white blood cells. The thymus is a primary lymphoid organ and the site of T cell maturation. Secondary lymphoid tissues (spleen and lymph nodes) are important as the site of lymphocyte (T and B cell) activation.

The lymphatic system and immunity

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Tonsils A collection of secondary lymphoid tissues in the throat. They provide defence against ingested or inhaled pathogens and produce activated B and T cells.

Thymus A primary lymphoid organ located above the heart. It is large in infants and shrinks after puberty to a fraction of its original size. Important for maturation of T-cells.

Spleen The largest mass of lymphatic tissue in the body. It stores and releases blood in case of demand (e.g. in severe bleeding), produces mature B-cells and antibodies and removes antibody-coated antigenic material.

Lymph nodes Ovoid masses of lymph tissue where lymphocytes are concentrated. Each node receives lymph through several narrow afferent (entry) vessels and exits via wider efferent (exit) vessels.

Red bone marrow A primary lymphoid tissue where all the different kinds of blood cells (including white blood cells) are produced by cellular differentiation from stem cells. B cells also mature here. Lymphatic vessels When the fluid leaking from capillaries is picked up by lymph capillaries, it is called lymph. The lymph, carrying leukocytes, flows in lymphatic vessels through the secondary lymphoid tissues.

The fluid circulating through the lymphatic system passes through the secondary lymphoid tissues, including the lymph nodes. These are ovoid organs, which are present throughout the lymphatic system. Lymph nodes receive lymph via incoming (afferent) vessels and are the site of lymphocyte activation. Lymphocytes in circulation are constantly moving between sites where antigens may be encountered. These antigens are presented to T cells in the secondary lymphoid tissues. Recognition of the antigen leads to activation and proliferation of both T and B cells, vastly increasing the number of lymphocytes. After several days, antigen-activated lymphocytes begin leaving the lymphoid tissue. Efferent lymph Bone marrow: immature lymphocytes

Bone marrow

Thymus

Lymph nodes and other secondary lymphoid tissue

Circulation of activated lymphocytes

Site of lymphocyte origin

Sites of maturation of B and T cells Sites of antigen presentation and activation of B and T cells

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1. What is the general role of the lymphatic system in immunity?

2. (a) What is the role of the secondary lymphoid tissue, e.g. lymph nodes, in the immune response?

(b) Why do you think lymph nodes become swollen when someone has an infection?

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81 The Adaptive Immune System are found in extracellular fluids including lymph, plasma, and mucus secretions and protect against viruses, and bacteria and their toxins. The cell-mediated immune response is associated with the production of specialised lymphocytes called T cells. Antigens are recognised by T cells only after antigen processing. The antigen is first engulfed by an antigen-presenting cell, which processes the antigen and presents it on its surface. T helper cells can then recognise the antigen and activate other cells of the immune system.

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Key Idea: Antigens, such as the cell walls of microbial cells, when processed by antigen-presenting cells, activate the B and T cells of the immune system against specific pathogens. There are two main components of the adaptive immune system: the humoral and the cell-mediated responses. They work separately and together to protect against disease. The humoral immune response is associated with the serum (the non-cellular part of the blood) and involves the action of antibodies secreted by B cells (B lymphocytes). Antibodies

Lymphocytes and their functions

Thymus gland

Bone marrow

B cells mature in the bone marrow in the shaft of the long bones (e.g. the femur). They migrate from here to the lymphatic organs.

The thymus gland is located above the heart. It is large in infants but regresses with age. Immature T cells move to the thymus to mature before migrating to other lymphatic organs.

Stem cell Stem cells in the bone marrow and fetal liver give rise to T cells and B cells.

B cell

T cell

Macrophage

T cell

Free antigen

Antigens

B cells recognise and bind antigens. Each B cell recognises one specific antigen. Helper T cells recognise specific antigens on B cell surfaces and induce their maturation and proliferation. A mature B cell may carry as many as 100,000 antigenic receptors embedded in its surface membrane. B cells defend against bacteria and viruses outside the cell and toxins produced by bacteria (free antigens).

Differentiate into two kinds of cells

T cells respond only to antigen fragments that have been processed and presented by infected cells or macrophages (phagocytic cells) (see opposite). They defend against: • Intracellular bacteria and viruses • Protozoa, fungi, flatworms, and roundworms • Cancerous cells and transplanted foreign tissue

Differentiate into various kinds of cells:

Antibody

TC

TH

Some B cells differentiate into long-lived memory cells (see Clonal Selection). When these cells encounter the same antigen again (even years or decades later), they rapidly differentiate into antibody-producing plasma cells.

When stimulated by an antigen (see Clonal Selection), some B cells differentiate into plasma cells, which secrete antibodies into the bloodstream. The antibodies then inactivate the circulating antigens.

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T cytotoxic cell destroys target cells on contact. Recognises tumour or virus-infected cells by their surface markers. Also called T killer cells.

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Plasma cells

There are also other types of T cells: T memory cells have encountered specific antigens before and can respond quickly and strongly when the same antigen is encountered again. T regulator cells control immune response by turning it off when no more antigen is present. They are important in the development of self tolerance. LINK

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Memory cells

T helper cell activates T cytotoxic cells and other helper T cells. They are needed for B cell activation.

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Dendritic cells stimulate the activation and proliferation of lymphocytes T helper cell

Dendritic cell

MHC receptor

B cell

T helper cell

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MHC II pathway

Presented antigen

Antigen

MHC I pathway

Cytokines (interleukin 12)

T killer cell

Dendritic cells, like macrophages, are antigen-presenting cells. Immature dendritic cells (DCs) originate in the bone marrow and migrate throughout the body. Once they have processed an antigen they begin to mature. They migrate to lymph nodes and, through antigen presentation and secretion of cytokines, stimulate the activation and proliferation of T cells. DCs exhibiting MHC I receptors stimulate the production of T cytotoxic cells. DCs exhibiting MHC II receptors stimulate the production of T helper cells. These in turn go on to stimulate the production of antibody-producing B cells.

1. Where do B cells and T cells originate (before maturing)? 2. (a) Where do B cells mature? (b) Where do T cells mature?

3. Describe the nature and general action of the two major divisions in the immune system:

(a) Humoral immune system:

(b) Cell-mediated immune system:

4. Explain how an antigen causes the activation and proliferation of T cells and B cells, including the role of dendritic cells:

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5. In what way do dendritic cells act as messengers between the innate and the adaptive immune systems?

(a) T helper cells:

(b) T cytotoxic cells:

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6. Describe the function of each of the following cells in the immune system response:

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82 Clonal Selection

Key Idea: Clonal selection theory explains how lymphocytes can respond to a large and unpredictable range of antigens. The clonal selection theory explains how the immune system can respond to the large and unpredictable range of potential antigens in the environment. The diagram below

Clonal selection theory Millions of B cells form during development. Antigen recognition is randomly generated, so collectively they can recognise many antigens, including those that have never been encountered. Each B cell has receptors on its surface for specific antigens and produces antibodies that correspond to these receptors. When a B cell encounters its antigen, it responds by proliferating and producing many clones that produce the same kind of antibody. This is called clonal selection because the antigen selects the B cells that will proliferate.

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Five (a-e) of the many B cells generated during development. Each one can recognise only one specific antigen.

describes clonal selection after antigen exposure for B cells. In the same way, a T cell stimulated by a specific antigen will multiply and develop into different types of T cells. Clonal selection and differentiation of lymphocytes provide the basis for immunological memory.

a

b

c

d

e

This B-cell encounters and binds an antigen. It is then stimulated to proliferate.

Memory cells

Plasma cells

Some B-cells differentiate into long lived memory cells.

The antibody produced corresponds to the antigenic receptors on the cell surface.

Some B-cells differentiate into plasma cells.

Antibodies are secreted into the blood by plasma cells where they inactivate antigens.

Some B cells differentiate into long lived memory cells. These are retained in the lymph nodes to provide future immunity (immunological memory). In the event of a second infection, memory B cells react more quickly and vigorously than the initial B cell reaction to the first infection.

Plasma cells secrete antibodies specific to the antigen that stimulated their development. Each plasma cell lives for only a few days, but can produce about 2000 antibody molecules per second. Note that during development, any B cells that react to the body’s own antigens are selectively destroyed in a process that leads to self tolerance (acceptance of the body’s own tissues).

1. Describe how clonal selection results in the proliferation of one particular B cell clone:

2. (a) What is the function of the plasma cells in the immune system response?

(b) What is the significance of B cells producing antibodies that correspond to (match) their antigenic receptors?

3. (a) Explain the basis of immunological memory:

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(b) Why are B memory cells able to respond so rapidly to an encounter with an antigen long after an initial infection?

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83 Antibodies

Key Idea: Antibodies are large, Y-shaped proteins, made by plasma cells, which destroy specific antigens. Antibodies and antigens play key roles in the response of the immune system. Antigens are foreign molecules which promote a specific immune response. Antigens include pathogenic microbes and their toxins, as well as substances such as pollen grains, blood cell surface molecules, and the Hinge region connecting the light and heavy chains. This allows the two chains to open and close.

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Variable regions form the antigen-binding sites. Each antibody can bind two antigen molecules.

surface proteins on transplanted tissues. Antibodies (or immunoglobulins) are proteins made in response to antigens. They are secreted from B cells into the plasma where they can recognise, bind to, and help destroy antigens. There are five classes of antibodies, each plays a different role in the immune response. Each type of antibody is specific to only one particular antigen.

Detail of antigen binding site

Antigen

Light chain (short)

Heavy chain (long)

Most of the molecule is made up of constant regions which are the same for all antibodies of the same class.

The antigen-binding sites differ from one type of antibody to another. The huge number of antibody types is possible only because most of the antibody structure is constant. The small variable portion is coded by a relatively small number of genes that rearrange randomly to produce an estimated 100 million different combinations.

Antibody

Symbolic form of antibody

Most antigens are proteins or large polysaccharides and are often parts of invading microbes. Examples include cell walls, flagella, toxins of bacteria, viral coats, and microbial surfaces.

The specific site on the antigen that is recognised by the immune system is called the epitope or antigenic determinant.

How antibodies inactivate antigens

Acting as agglutinins

Acting as antitoxins

Enhancing phagocytosis Phagocyte

Antibody

Chemical marker

Soluble antigens

Toxins

Antigen/bacteria

Antibodies can act as agglutinins and cause antigens to bind together, forming inactivated clumps.

Antibodies can act as antitoxins by binding to toxins and neutralising them.

Tags foreign cells for destruction by phagocytes.

1. Describe the structure of an antibody, identifying the specific features of its structure that contribute to its function:

(a) Acting as agglutinins:

(b) Acting as antitoxins:

(c) Working with opsonins:

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2. Explain how the following actions by antibodies enhance the immune system's ability to stop infections:

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84 KEY TERMS AND IDEAS: Did You Get It?

1. Match each term to its definition, as identified by its preceding letter code.

adaptive immune response antibodies

encountered in the past.

B Lymphocytes that make antibodies against specific antigens. C Large phagocytic and amoeboid white blood cells within tissues with critical roles in both innate and adaptive immune responses.

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antigen

A The ability of the immune system to respond rapidly in the future to antigens

B cells

cell mediated immunity clonal selection

D Generalised defence mechanisms against pathogens, e.g. physical barriers, secretions, inflammation, and phagocytosis.

E A molecule that is recognised by the immune system as foreign.

F An antigen-specific immune response with both cell-mediated and humoral components. Characterised by immunological memory.

dendritic cell immunity

immunological memory

G A model for how B and T cells are selected to target specific antigens invading the body.

H Secondary lymphoid tissues important as the site for lymphocyte activation and proliferation in response to antigens.

I White blood cells that destroy foreign material, e.g. bacteria, by ingesting them.

innate immunity lymph nodes lymphocytes

macrophage phagocytes

J Immune response against antigens involving the activation of macrophages, specific T cells, and cytokines.

K Gamma globulin proteins in the blood or other bodily fluids, which identify and neutralise foreign material, such as bacteria and viruses.

L Specific white blood cells involved in the adaptive immune response.

M The collective defences of the body that provide resistance to infection or disease. N A type of white blood cell that processes antigen material and presents it to T cells.

2. Contrast the innate and the adaptive immune responses with reference to the basic action and the cells involved:

3. The photograph on the right shows the effect of a pathogen infecting a human.

(b) What is happening to the blood vessels at this location?

(c) Name the substance responsible for the change in the blood vessels:

(d) What type of cell is the substance released from?

(e) During this response, the number of white blood cells increases/decreases (delete one).

(f) The process occurring here is an example of innate immunity / adaptive immunity (delete one).

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(a) Name the defensive response occurring:

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Immunity

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Unit 3 Outcome 2

Key terms

Natural and artificial immunity

active immunity

Key knowledge

allergen

allergic reaction antigen

c

1

c

2

Distinguish between naturally acquired and artificially acquired immunity.

85

Describe active and passive strategies for acquiring immunity.

85

artificially acquired immunity

Vaccination

asthma

Key knowledge

autoimmune disease

3

Explain what is meant by herd immunity and explain its importance to the resistance of the population to circulating pathogens.

86

c

4

Explain the principles of vaccination, including reference to the primary and secondary response to infection and the role of these.

85

c

5

Explain the role of vaccination programmes in maintaining herd immunity for a particular disease in the population.

histamine

hypersensitivity

Activity number

c

hayfever

herd immunity

Activity number

immune deficiency disease

86-87

immunity

immunised

immunological memory

monoclonal antibody multiple sclerosis

naturally acquired immunity

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passive immunity

Immune system diseases

primary response

c

6

vaccination

With reference to specific examples, describe deficiencies and malfunctions of the immune system. Include reference to: (a) Autoimmune diseases (e.g. multiple sclerosis). (b) Immune deficiency diseases (e.g. HIV/AIDS). (c) Allergic reactions and hypersensitivity (e.g. asthma and hayfever).

Monoclonal antibodies

91 92

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Describe the applications of monoclonal antibodies in treating cancer, including reference to principles and mode of action.

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88 89 90

Activity number

Key knowledge

c

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Activity number

Key knowledge

secondary response sensitisation

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85 Acquired Immunity

Key Idea: Acquired immunity is a resistance to specific pathogens acquired over the life-time of an organism. We are born with natural or innate resistance which provides non-specific immunity to certain illnesses. In contrast, acquired immunity is protection developed over time to specific antigens. Active immunity develops after

the immune system responds to being exposed to microbes or foreign substances. Passive immunity is acquired when antibodies are transferred from one person to another. Immunity may also be naturally acquired, through natural exposure to microbes, or artificially acquired as a result of medical treatment (below).

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Acquired immunity

Naturally acquired

Artificially acquired

Active

Passive

Active

Passive

Antigens enter the body naturally, as in cases where: • Microbes cause the person to actually catch the disease • Sub-clinical infections (those that produce no evident symptoms). The body produces antibodies and specialised lymphocytes.

Antibodies pass from the mother to the fetus via the placenta during pregnancy, or to her infant through her milk. The infant's body does not produce any antibodies of its own.

Antigens (non-infectious parts of microbes) are introduced in vaccines (a process called vaccination). The body produces antibodies and specialised lymphocytes and becomes immunised.

Pre-formed antibodies in an immune serum are introduced into the body by injection (e.g. antivenom used to treat snake bites). The body does not produce antibodies.

1. (a) What is meant by passive immunity?

(b) Distinguish between naturally and artificially acquired passive immunity and give an example of each:

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2. (a) Why does a newborn baby need to have received a supply of maternal antibodies prior to birth?

(b) Why is this supply supplemented by antibodies in breast milk?

(c) Would you recommend breast feeding to a new mother? Explain your answer:

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If a person has not been immunised against a disease, exposure to the pathogen causes them to become ill and their body forms antibodies against it.

Antibodies passing from the mother's milk to her newborn baby provide protection until the baby develops its own antibodies.

Vaccines provide immunity to specific pathogens and greatly reduce the risk of contracting the disease.

Without treatment with the appropriate preformed antibodies in antivenom, the bites of many snake species can be fatal.

When the B cells encounter antigens and produce antibodies, the body develops active immunity against that antigen.

The initial response to antigenic stimulation, caused by the sudden increase in B cell clones, is called the primary response. Antibody levels as a result of the primary response peak a few weeks after the response begins and then decline. However, because the immune system develops an immunological memory of that antigen, it responds much more quickly and strongly when presented with the same antigen subsequently (the secondary response). This forms the basis of immunisation programmes where one or more booster shots are provided following the initial vaccination.

Amount of antibody in the serum (arbitrary units)

Primary and secondary response to antigens Primary and secondary response to antigens

Secondary response

First antigen injection

Second antigen injection

Primary response

0

40

80

120

160

200

240

280

Time after administration of antigen (days)

3. (a) What is active immunity?

(b) Distinguish between naturally and artificially acquired active immunity and give an example of each:

(b) Why is the secondary response so different from the primary response?

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4. (a) Describe two differences between the primary and secondary responses to presentation of an antigen:

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86 Vaccines and Vaccination produce antibodies against the antigen, but it does not cause the disease. The immune system remembers its response and will produce the same antibodies if it encounters the antigen again. If enough of the population are vaccinated, herd immunity (indirect protection) provides unvaccinated individuals in the population with a measure of protection against the disease. There are two basic types of vaccine, subunit vaccines and whole-agent vaccines (below).

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Key Idea: A vaccine is a suspension of antigens that is deliberately introduced into the body to protect against disease. If enough of the population are vaccinated, herd immunity provides protection to unvaccinated individuals. A vaccine is a preparation of a harmless foreign antigen that is deliberately introduced into the body to protect against a specific disease. The antigen in the vaccine is usually some part of the pathogen and it triggers the immune system to

Types of Vaccine

Whole-agent vaccine Contains whole, nonvirulent microorganisms

Subunit vaccine

Inactivated (killed)

Attenuated (weakened)

Viruses for vaccines may be inactivated with formalin or other chemicals. They present no risk of infection, e.g. most influenza vaccines, Salk polio vaccine.

Attenuated viruses are usually strains in which mutations have accumulated during culture. These live viruses can back-mutate to a virulent form, e.g. MMR vaccine.

Contains some part or product of microbes that can produce an immune response. Includes vaccines made using genetic engineering, inactivated toxins, and conjugated and acellular vaccines, e.g. the diphtheria-tetanus-pertussis vaccine and the vaccine against bacterial meningitis.

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Why are vaccinations given?

1. (a) What is a vaccine?

(b) Provide some examples of when vaccinations are needed:

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Tourists may need specific vaccines if the country they are visiting has a high incidence of a certain disease. For example, travellers to South America should be immunised against yellow fever, a disease that does not occur in Australia.

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Most vaccinations are given in childhood, but adults may be vaccinated against a disease (e.g. TB, tetanus) if they are in a high risk group (e.g. the elderly or farmers) or to provide protection against seasonal diseases such as influenza.

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Vaccines against common diseases are given at various stages during childhood according to an immunisation schedule. Vaccination has been behind the decline of some once-common childhood diseases, such as mumps and measles.

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Vaccination can provide herd immunity Herd immunity occurs when the vaccination of a significant portion of a population provides some protection for individuals who have not developed immunity (e.g. have not been vaccinated and are not immunised). In order to be effective for any particular disease, a high percentage of the population needs to be vaccinated against that disease. High vaccination rates make it difficult for the disease to spread because there are very few susceptible people in the population. Herd immunity is important for people who cannot be vaccinated (e.g. the very young, people with immune system disorders, or people who are very sick, such as cancer patients).

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Immunised and healthy

Not immunised and healthy

Not immunised, sick and contagious

The level of vaccination coverage to obtain herd immunity differs for each disease. Highly contagious diseases (e.g. measles) need a much higher vaccine uptake (95%) than a less contagious disease such as polio (80-85%).

High herd immunity: Most of the population is immunised. The spread of the disease is limited. Only a few people are susceptible and become infected.

Low herd immunity: Only a small proportion of the population is immunised. The disease spreads more readily through the population infecting many more people.

2. Attenuated viruses provide long term immunity to their recipients and generally do not require booster shots. Why do you think attenuated viruses provide such effective long-term immunity when inactivated viruses do not?

3. (a) What is herd immunity?

(b) Why are health authorities concerned when the vaccination rates for an infectious disease fall?

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4. Some members of the population are unable to be vaccinated. Give an example and explain why herd immunity is very important to them?

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87 Vaccines Can Eliminate Infectious Disease

Key Idea: Vaccination programmes have been successful in the global eradication of smallpox, but other diseases (such as influenza) are more difficult to eradicate. To date, the only infectious disease globally eradicated has been smallpox. Several factors lead to this success. Smallpox is easily identifiable by its characteristic rash making surveillance and containment of infected patients

easier. It has no other natural carriers so once immunisation rates reached a critical level, its spread through the population was limited. Other diseases can be more difficult to eradicate. This is especially true for diseases that have a long period between infection and the symptoms showing (e.g. TB) or diseases caused by pathogens with high rates of mutation (e.g. influenzavirus or HIV).

The challenges of eradicating disease

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Measles elimination in Australia

Measles vaccine added to schedule

800 700

Measles-mumps vaccine MMR (12 months)

600 500 400

MMR (10-16 yrs)

300 200

1995

1990

1985

1980

1975

1970

0

2000

MCC

100

Year MMR: measles-mumps rubella vaccine MCC: measles control campaign The graph above shows the role of vaccination in reducing

measles hospitalisations in the state of Victoria. MMR is the introduction of the measles/mumps/rubella vaccine. MCC (measles control campaign) was an extensive mass vaccination and monitoring campaign.

A child with whooping cough

CDC

900

Data source: Bulletin of the World Health Organisation, 2008

Measles-associated hospitalisations in Victoria, Australia (1962-2004)

1965

Number of hospitalised cases

Measles is a highly contagious disease, one infected person can infect 12-18 people during their infectious period. In 2014, the World Health Organisation (WHO) announced measles had been eliminated from Australia. High vaccination rates contributed to its elimination. However, measles still occurs in other countries so it could be reintroduced if an infected traveller entered Australia. Maintaining high levels of vaccination is important in preventing its reintroduction.

Whooping cough is a respiratory disease caused by the bacterium Bordetella pertussis. Despite high vaccination rates, whooping cough is increasing in Australia. Several factors may be contributing to this.

ffUntil 1997, a whole vaccine was used. It

contained hundreds of different antigens and provided protection against many strains of the pertussis pathogen. In 1999, an acellular vaccine (right) was introduced. It only contains five antigens and so provides less protection.

ffNew strains of B. pertussis are evolving, and the new vaccine is not effective against them.

ffMore adults who were vaccinated

against whooping cough in childhood are contracting the disease. This suggests the effectiveness of the vaccine declines over time.

1. Study the measles graph above and describe the role of vaccination programmes in eliminating measles from Australia:

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2. What could happen if vaccination rates for measles fell too low?

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3. Give a likely explanation for why whooping cough rates are increasing in Australia despite high vaccination rates:

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88 Autoimmune Disease: Multiple Sclerosis

Key Idea: Autoimmune diseases are the result of the body's immune system attacking its own tissues. Any of numerous disorders, including rheumatoid arthritis, type 1 diabetes mellitus, and multiple sclerosis, are caused by an immune system reaction to the body's own tissues. The immune system normally distinguishes self from non-self. Autoimmune diseases occur when this normal recognition system fails and a certain cell or tissue type

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T cells incorrectly recognise the myelin sheath as a foreign antigen and attack it.

is no longer recognised as self. The exact mechanisms behind autoimmune malfunctions are not fully understood, but pathogens or drugs may play a role in triggering an autoimmune response in someone who already has a genetic predisposition. The reactions are similar to those that occur in allergies, except that in autoimmune disorders, the hypersensitivity response is to the body itself, rather than to an outside substance.

Macrophage

Inflammation recruits more immune cells (e.g. macrophages, B cells, and T cells) to the site and more damage to the myelin occurs.

B cells secrete myelin reactive antibodies.

T cell releases cytokines, causing inflammation.

Nerve cell

Myelin is progressively destroyed.

Multiple sclerosis (MS) is the most common autoimmune disease affecting the central nervous system. MS is a progressive inflammatory disease in which scattered patches of myelin in the brain and spinal cord are destroyed. Myelin (left) is the protective fatty connective tissue sheath surrounding conducting axons (photo left). Its main function is to insulate the axon and so increase the speed at which nerve impulses are conducted. Its destruction disrupts transmission of electrical impulses and leads to neural impairment. Once the myelin is damaged, the unprotected axon can also become scarred and damaged by continued immune system attack.

Myelin

Axon

Roadnottaken

The underlying axon can become scarred and damaged.

MS usually starts early in adult life. There is a genetic component to the disease, as relatives of affected people are eight times more likely to contract the disease. Symptoms include numbness, tingling, muscle weakness and paralysis. There is currently no cure.

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1. What is an autoimmune disease?

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2. Using the example of multiple sclerosis, explain how immune system malfunction can result in autoimmune disease:

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89 HIV/AIDS: An Immune Deficiency Disease

Key Idea: The human immunodeficiency virus (HIV) infects lymphocyte cells, eventually causing AIDS, a fatal disease, which acts by impairing immune system function. Immune deficiency occurs when the body's immune system has limited (or no) ability to fight infectious disease. People with immune deficiency diseases become sick more often and for longer periods than a healthy individual. Some types of immune deficiency are inherited, but others develop as a

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result of another factor (e.g. chemotherapy, malnutrition, or infectious disease). HIV (human immunodeficiency virus) causes immune deficiency by destroying T helper cells, which are central to cellular immunity. Over time, a disease called AIDS (acquired immunodeficiency syndrome) develops and the immune system loses its ability to fight off infections as more T helper cells are destroyed. There is no cure or vaccine for HIV, but some drugs can slow the progress of the disease.

HIV infects T helper cells. It uses the cells to replicate itself in great numbers, then the newly formed viral particles exit the cell to infect more T helper cells. Many T helper cells are destroyed by the viral replication. Because of their role in cellular immunity, T helper cell destruction recruits more T cells, accelerating the infection of new cells.

HIV budding from a lymphocyte

900

Category B Some symptoms, low T helper cell count

Category C Clinical AIDS symptoms appear

800 700

T helper cell population

600 500 400 300

HIV population

200 100

CDC

HIV

Category A HIV positive with few or no symptoms

0

0

1

2

3

4

5 Years

6

7

8

9

106 105 104 103 102

Viral RNA (copies per mL)

Once the T helper cell population becomes depleted, the immune system's ability to fight infection is severely compromised.

The graph below shows the relationship between the level of HIV infection and the number of T helper cells in an individual.

T helper cell concentration in blood (cells per ÂľL)

HIV infects T helper cells

101 0 10

HIV uses the cellular machinery of T helper cells to replicate HIV hijacks the T helper cells' machinery to replicate itself. Reverse transcriptase produces double stranded DNA (dsDNA) from the viral RNA. The host cell transcribes the viral genes to produce new viruses.

Viral RNA

dsDNA

Viral genes expressed

HIV binds to specific CD4 receptors on T helper cells. HIV fuses with the plasma membrane of the T helper cell. Its RNA enters the cell.

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The HIV particles mature and infect more T helper cells. As more T helper cells become infected, the body's immune response weakens.

T helper cell

Assembled viruses exit cell and reinfect

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HIV

The new HIV particles bud from the T helper cell. Between 1000 and 3000 new HIV particles can be released from a single infected cell.

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The genetic material of HIV is a single strand of RNA.

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AIDS: The end stage of an HIV infection HIV/AIDS is a spectrum of disorders (right) arising as a consequence of impaired immune function, which prevents the body detecting and destroying pathogens or damaged cells.

Dermatitis especially on the face.

A variety of opportunistic infections, including Herpes and tuberculosis. Oral thrush affecting respiratory tract.

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People with healthy immune systems can fight off the challenges of pathogens and are able to detect and destroy damaged (pre-cancerous) cells. However, people with HIV are susceptible to all pathogens because their resistance to disease is so low. Moreover, loss of the T cell population compromises the ability of HIV-infected people to detect and destroy pre-cancerous cells. Rare cancers are a common symptom of HIV/AIDS.

Eye infections (Cytomegalovirus)

Fever, cancer, toxoplasmosis of the brain, and dementia.

Antibiotics can be used to treat some of the infections contracted due to the reduced immune system (e.g. tuberculosis), but they cannot be used to treat the HIV infection itself. Antibiotics work by targeting prokaryotic (bacterial) metabolism and so are ineffective against a non-cellular pathogen that hijacks eukaryotic metabolism in order to replicate. Although there is currently no cure of HIV/AIDS, some drugs can slow the progress of the disease by interfering with the replication of HIV and slowing the advance of the disease.

Kaposi's sarcoma: a highly aggressive malignant skin tumour. Usually starts at the feet and ankles, spreading throughout the body.

Marked weight loss and infectious diarrhea. A number of autoimmune diseases, especially destruction of platelets.

1. (a) What type of cells does HIV infect?

(b) How does HIV recognise this type of cell?

(c) What is the role of reverse transcriptase in HIV replication?

2. Study the graph on the previous page showing how HIV affects the number of T helper cells. Describe how the viral population changes with the progression of the disease:

3. (a) What effect does HIV have the cells of the immune system?

4. (a) Why is the purpose of antibiotics in treatment of HIV/AIDS?

(b) Why are antibiotics ineffective against the HIV infection itself?

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(b) Describe the effect of this change on the long-term health of a person with HIV:

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90 Allergies and Hypersensitivity phases. The first is sensitivity, in which the body becomes primed to respond to an allergen but there are no symptoms. This is followed by a reaction phase in which the person is re-exposed to the same allergen and experiences symptoms. In some cases, the response causes only minor discomfort (e.g. hayfever) but in extreme cases it can cause death.

Sensitisation phase

Reaction phase

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Key Idea: Hypersensitivity occurs when the immune system overreacts to an antigen or reacts to the wrong substance. Histamine plays a significant role in this response. Sometimes the immune system may overreact, or react to the wrong substances instead of responding appropriately. This response is called an allergy, and it consists of two

A

Dendritic cell

The same allergen to which a person is already sensitised encounters a mast cell. This second encounter can be weeks or months after the initial encounter.

Pollen allergen

MHC II receptor

Activated T helper cell

The allergen binds to specific IgE antibodies on the mast cell. Once bound, the allergen becomes cross-linked to neighbouring antibodies.

B

Inactivated T cell

Antibody

Several chemicals, including histamine are released from the mast cell. Several reactions occur in response to histamine including dilation of blood vessels, inflammation, and mucus secretion.

C

B cell

Mast cell

When a new allergen is encountered, it is phagocytosed by a dendritic cell and digested into smaller antigenic fragments. The fragments are displayed by the dendritic cell on MHC class II receptors. When an inactivated (or naive) T cell encounters the novel allergen it becomes an activated T helper cell. Cytokines called interleukins produced by the T helper cell activate B cells to produce IgE antibodies. IgE antibodies bind to the surface of mast cells where they remain. No allergic symptoms occur.

Hayfever is a common allergic reaction to airborne allergens such as dust, moulds, pollen, animal fur, and feathers. Histamine causes the symptoms associated with hayfever including wheezing, inflammation, itching, sneezing, and watering of the eyes and nose.

Eyewire

Histamine

2. What is the role of histamine in hypersensitivity responses?

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1. What happens when a person becomes sensitised to an allergen?

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3. In what way is the hypersensitivity reaction a malfunction of the immune system?

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91 Monoclonal Antibodies useful because they are identical (i.e. clones), they can be produced in large quantities, and they are highly specific for a particular antigen. Most monoclonal antibodies are produced in mice, and in some people the foreign mouse proteins can cause an unwanted immune response. Monoclonal antibodies have wide applications in diagnosing and treating disease, in detecting pregnancy, and in food safety tests.

Culture of tumour cells (mutant myeloma cells). These cells are immortal and do not stop dividing.

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Key Idea: Monoclonal antibodies are artificially produced antibodies that neutralise specific antigens. They have many diagnostic and therapeutic applications. A monoclonal antibody is an artificially produced antibody that binds to and neutralises one specific type of antigen. A monoclonal antibody binds an antigen in the same way that a normally produced antibody does. Monoclonal antibodies are

Making monoclonal antibodies The mouse’s B-lymphocyte cells have developed an antibody to recognise the foreign protein (antigen).

A mouse is injected with a foreign protein (antigen) that will stimulate the mouse to produce antibodies against it.

Pure tumour cells are harvested

A few days later, B-lymphocytes (that make the antibodies) are taken from the mouse’s spleen.

Ethical issues with the use of monoclonal antibodies

The main ethical arguments against the use of monoclonal antibodies include:

ffAnimal testing: Most antibodies

are produced and harvested from mice. However new in vitro techniques are in development where no animals will be required to produce monoclonal antibodies.

Mouse cell and tumour cell fusing

The mouse cells and tumour cells are mixed together in suspension Unfused cells also present

Hybridoma cell

Some of the mouse cells fuse with tumour cells to make hybrid cells called hybridomas.

ffGenetic engineering: To make

The mixture of cells is placed in a selective medium that allows only hybrid cells to grow.

monoclonal antibodies stable for use in humans, human genes are transplanted into mice.

ffHuman safety: When used, some

monoclonal antibodies have caused deaths in recipients.

Hybrid cells are screened for the production of the desired antibody. They are then cultured to produce large numbers of monoclonal antibodies.

1. (a) What is a monoclonal antibody?

(b) In what way are monoclonal antibodies the same as a regular antibody?

2. (a) Which mouse cells are used to produce monoclonal antibodies?

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(b) What potential health problem is associated with the use of mice to produce monoclonal antibodies?

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3. Which characteristic of tumour cells allows an ongoing culture of antibody-producing lymphocytes to be made?

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What is ELISA?

Applications of monoclonal antibodies Diagnostic uses • Detecting the presence of pathogens, distinguishing between viral strains, and diagnosing HIV/AIDS. • Measuring protein, toxin, or drug levels in serum. • Blood and tissue typing. • Detection of antibiotic residues in milk.

Therapeutic uses

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ELISA stands for enzyme-linked immunosorbent assay. ELISA is a technique used to detect and quantify a specific antigen (e.g. peptides, proteins, and hormones). Antigens from the sample being tested are attached to a surface, such as a microtiter plate (below). An antibody specific to the antigen is applied and binds to the antigen. The antibody has an enzyme attached. Lastly, a solution containing the enzyme's substrate is added. The reaction between the enzyme and its substrate produces a colour change in the substrate. The colour change indicates how much antigen is present in the sample. Several variations in ELISA methods exist. Pregnancy testing is an example of use of ELISA.

• Neutralising bacterial endotoxins (e.g. in blood infections). • Preventing organ rejection, e.g. in kidney transplants, by interfering with the T-cells involved in the rejection of transplanted tissue.

• Treatment of some hypersensitivity disorders such as asthma. The monoclonal antibodies inhibit IgE antibodies binding to mast cell receptors. This limits the inflammatory response and therefore the extent of the reaction. • Detection and treatment of some cancers. Herceptin is a monoclonal antibody used to treat breast cancer. Herceptin recognises receptor proteins on the outside of cancer cells and binds to them. The immune system can then identify the antibodies as foreign and destroy the cell.

A well that undergoes a dark colour indicates more antigen is present than in a sample with a lighter colour change.

• Inhibition of platelet clumping in patients who have had their coronary artery surgically unblocked. The monoclonal antibodies bind to the receptors on the platelet surface preventing them from clotting.

4. (a) Describe some diagnostic uses of monoclonal antibodies:

(b) What is a likely target for a monoclonal antibody used in the diagnosis of HIV?

5. What role do monoclonal antibodies have in treating hypersensitivity disorders such as asthma?

6. (a) What is the purpose of ELISA?

(b) An ELISA set up is depicted on the right. What does each labelled component represent?

B: C:

C

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7. Discuss some of the ethical issues associated with the use of monoclonal antibodies:

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92 Herceptin: A Modern Monoclonal

Key Idea: Herceptin is a monoclonal antibody that attaches to the HER2 receptor protein on cells to help T-cells target them for destruction. Herceptin is the patented name of a monoclonal antibody for the treatment of breast cancer. It targets the HER2 receptor proteins on cancerous cells that signal to the cell when it should divide. The proteins are produced by the protooncogene HER2. Cancerous cells contain 20-30% more

of the HER2 gene than normal cells and this causes overexpression of HER2, and large amounts of HER2 protein. The over-expression causes the cell to divide more often than normal, producing a tumour. HER2 protein is not a foreign protein, so cancerous (HER2+) cells are not recognised by the immune system as abnormal. Herceptin binds to the HER2 protein on the surface of the cancerous cell so that the immune system can recognise it as foreign and destroy it.

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Herceptin targeted destruction of cancer cells

HER2 gene (Human Epidermal growth factor Receptor 2) produces proteins that migrate to the surface of the affected cell.

The cancerous cell over-expresses the HER2 gene, producing large amounts of the HER2 protein on the surface of the cell.

HER2 protein on cell surface

Over-expression of HER2 gene

HER2 gene in nucleus of cell

Receptor site for Herceptin

Receptor sites on the T-cells recognise the Herceptin antibody and attach to it.

Herceptin (monoclonal antibody) attached to receptor protein.

T-cells recognise the antibody as foreign and attack it, destroying the cell and reducing the tumour growth.

T-cell of the immune system destroys cells directly

Herceptin recognises the HER2 protein and attaches to it.

1. (a) Why does the immune system not detect HER2+ cells as abnormal and destroy them?

(c) Study the graph (right). What effect does Herceptin have on survival rates of women treated for HER2+?

100 90 80

Chemotherapy only

70

Chemotherapy + Herceptin

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cells?

Effect of herceptin on survival rate in HER2+ breast cancer patients

1

2 3 4 Years after randomisation

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(b) How does Herceptin detect and destroy

Overall survival (percent)

HER2+

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93 KEY TERMS AND IDEAS: Did You Get It?

1. Test your vocabulary by matching each term to its correct definition, as identified by its preceding letter code. active immunity

A The status of a person with resistance against a particular pathogen as a result of vaccination.

allergen

allergic reaction

has encountered before.

C Any substance that causes an allergic reaction.

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artificially acquired immunity

B The more rapid and stronger response of the immune system to an antigen that it

D Antibodies made by clones of one type of immune cell.

autoimmune disease

E Disorders in which immune system function is impaired, often because of loss of

herd immunity

F Long-lasting immunity that is induced in the host itself by the antigen.

histamine

particular types of immune cells.

G Immunity developed through artificial exposure to antigens (e.g. vaccination).

H The delivery of antigenic material to produce immunity to a disease (produce an immunised individual).

hypersensitivity

immune deficiency disease immunised

I An autoimmune disorder that results in destruction of myelin around nerve cells and results in neural impairment.

J The initial response of the immune system to exposure to an antigen.

K Indirect protection from an infectious disease provided to susceptible individuals

because most of the population has been immunised against a specific pathogen.

monoclonal antibodies multiple sclerosis

naturally acquired immunity

L The first phase of an allergic response in which the body becomes primed to respond to an allergen but there are no symptoms.

M Immunity developed through the natural exposure to antigens (e.g. natural exposure to a pathogen).

passive immunity

N The hypersensitive response of the immune system in response to an allergen it

primary response

O A disease resulting from the inappropriate targeting and destruction of the body's

has previously been sensitised to. own cells and tissues.

secondary response

P A chemical substance released by mast cells and involved in inflammatory and allergic responses.

sensitisation vaccination

Q An undesirable overreaction of a normal immune system to antigens. R Immunity gained by the receipt of ready-made antibodies.

2. The diagram below shows a hypersensitivity sequence. Label the following components on the diagram: mast cell, antibody, histamine, antigen, B cell. In the space below each drawing, briefly state what is happening.

4

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94 Review: Unit 3, Area of Study 2

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Summarise what you know about this topic under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts. Use the images and hints to help you and refer back to the introduction to check the points covered: Cellular signals

Responding to antigens

HINT: Give examples of types of antigens. Distinguish between the innate and adaptive immune responses.

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HINT: Describe signal transduction and types of signalling molecules. Outline the process and roles of apoptosis.

Immunity

REVISE

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HINT: What is immunity and how can it be invoked? Give example of immune system malfunctioning. What role can monoclonal antibodies play in treating cancer?

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95 Synoptic Assessment: Unit 3, Area of Study 2

1. Cells use signals (chemical messengers) to communicate and to gather information about, and respond to, changes in their cellular environment. The diagram below shows a form of cell signalling. (a) Describe the basis of cell signalling in terms of a receptor, cell signalling molecule, and a target cell:

A

Extracellular fluid

Plasma membrane

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(b) Why is it important that not all cells respond to every cell signal?

Cytoplasm

B

C

(c) On the diagram, what type of signal molecule is A? Explain your reasoning:

(d) What is the structure labelled B?

D

(e) What role is structure B playing at point C on the diagram?

(f) What is happening at point D on the diagram?

(g) In general terms, describe the effect of this on the cell:

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2. Discuss the components and the importance of the innate immune response in the initial defence against a pathogen:

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3. Discuss how the adaptive immune system responds to a pathogen and how this initial exposure can lead to a quicker response following a second exposure. Explain how this phenomenon is used in vaccination against disease.

4. Pertussis, commonly known as whooping cough, is a highly contagious respiratory infection caused by the bacterium Bordetella pertussis. In Australia, the vaccine is given at two, four and six months of age. Booster doses are given at 18 months, four years and between 10-15 years. Epidemics occur in Australia every 3-4 years, but between 2008-2012 there has been a significant increase in whooping cough. Most of the people contacting whooping cough have been adults who had been immunised in childhood. (a) Suggest why many adults are contracting whooping cough (even though they were vaccinated as children):

(b) What effect could the high number of whooping cough cases have on young children who have not yet completed their vaccination schedule?

(c) How could the rates of whooping cough be reduced in Australia?

(d) The vaccine for whooping cough is an acellular vaccine. Discuss the advantages and disadvantages of this:

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(a) What type of immunity is this?

(b) What has occurred at point A?

(c) What has occurred at point B?

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Amount of antibody in the serum (arbitrary units)

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5. The graph right shows a primary and secondary immune response to an artificially introduced antigen.

B

A

(d) Describe and explain the differences in the amount of antibody detected after each of the two events:

Time after administration of antigen

Measles

1400

% vaccination

1200

Band for effective herd immunity 83-94%

1000

95 90

800 600

Wakefield’s publication

400 200

0 1994

1996

1998

2000

2002 Year

2004

2006

2008

85

80

75 2010

The graph above shows the number of measles cases in the UK, together with percentage vaccination, 1994-2008.

(a) What happened to MMR vaccination rates after the publication of Wakefield's study?

(b) What is the trend in measles cases in the UK since 2006?

(c) Give a likely explanation for this trend:

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Percentage vaccinated

UK measles cases and vaccination

1600

Notified cases

6. In 1998, Dr Andrew Wakefield and his colleagues published a paper linking the measles, mumps, and rubella vaccine (MMR) to an increase in autism rates. As a result, the uptake of the MMR vaccine in the UK dropped, and several measles outbreaks occurred. Dr Wakefield's paper has since been retracted by the journal in which it was published as it was found to be fraudulent and flawed in several aspects, e.g. sample size of only 12, with no control group. Since the publication of Wakefield's paper, 20 large scale epidemiologic studies into MMR and autism have been carried out in several countries. All have shown that the MMR vaccine does not cause autism. However, the damage has been done, and health authorities must now convince the public that the vaccine is safe.


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Changes in the genetics of populations

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Unit 4 Outcome 1

Key terms

Causes of changing allele frequencies

adaptation

Key knowledge

allele

allele frequency

1

Understand the concept of the gene pool and explain how allele frequencies are expressed for populations. Recognise evolution as the change in the allele frequency of a population over time.

c

2

Explain the effect of mutation, gene flow (migration), natural selection, and genetic drift on the allele frequencies of populations.

96 97 109 110

c

3

Describe gene and chromosome mutations as a source of new alleles, including point mutations, frame shifts, and block (chromosome) mutations.

98 - 101 104 106

c

4

Describe chromosomal abnormalities (aneuploidies and polyploidies) including reference to how they arise and their biological consequences.

102 103

c

5

Explain the mechanism of natural selection, knowing that it acts on phenotypic variation in the population. Understand the role of natural selection in sorting the variability within a gene pool and establishing adaptive genotypes.

107

c

6

Understand what is meant by (evolutionary) fitness. Explain how evolution, through adaptation, equips species for survival.

108

c

7

Describe examples of stabilising, disruptive, and directional selection. How are deleterious alleles maintained in the population? Why might they not disappear?

c

8

differential survival

directional selection disruptive selection evolution fitness

fixation (of alleles) founder effect gene flow

gene pool

genetic bottleneck genetic drift

genetic equilibrium genotype

heterozygous advantage

97

c

allopatric speciation aneuploidy

Activity number

c

9

c

10

111 - 121

Explain the genetic and evolutionary consequences of the founder effect.

122

Explain the genetic and evolutionary consequences of the bottleneck effect.

123

Recognise genetic drift as an important process in evolution. Describe its consequences and the conditions under which it is important (see #8 and 9).

124

mate choice mutation

natural selection phenotype polyploidy

population

postzygotic isolating mechanism

Prof. Jeff Podos

prezygotic isolating mechanism

Allopatric speciation

reproductive isolation

Key knowledge

Menna Jones

Activity number

c

11

Describe and explain mechanisms of reproductive isolation, distinguishing between prezygotic and postzygotic reproductive isolating mechanisms.

125 126

selective breeding (=artificial selection)

c

12

Explain allopatric speciation, including reference to how mutation and different selection pressures acting on isolated populations lead to reproductive isolation.

126 - 128

speciation stabilising selection

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selection pressure

Manipulating gene pools through selective breeding Key knowledge

Activity number

13

Explain how selective breeding (artificial selection) can alter the allele frequencies of a gene pool and cause phenotypic and genotypic change.

129-133

c

14

Describe examples to show how selective breeding has created phenotypic and genotypic change in populations. Examples include the development of modern domestic livestock breeds and crop varieties from their wild ancestors.

129-133

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96 Sources of Variation combinations) and phenotypes (appearances) in a population. Variation in phenotypic characteristics, such as flower colour and birth weight, is a feature of sexually reproducing populations. Some characteristics show discontinuous variation, with only a limited number of phenotypic variants in the population. Others show continuous variation, with a range of phenotypic variants approximating a bell shaped (normal) curve. Both genotype and the environment determine, to different degrees, the final phenotype we see.

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Key Idea: The characteristics of sexually reproducing organisms show variation. Those showing continuous variation are controlled by many genes at different loci and are often greatly influenced by environment. Those showing discontinuous variation are controlled by a small number of genes and there are a limited number of phenotypic variants in the population. Both genes and environment contribute to the final phenotype on which natural selection acts. Variation refers to the diversity of genotypes (allele

Mutations

Sexual reproduction

gene mutations, chromosome mutations Mutations are the source of all new alleles. Existing genes are modified by mutations to form new alleles. Neutral mutations, which are neither detrimental nor beneficial, may escape selection pressure until conditions change.

independent assortment, crossing over and recombination, mate selection Sexual reproduction rearranges and reshuffles the genetic material into new combinations in the offspring.

Mutation: T instead of C

Mutated DNA

Phenotype

An individual’s phenotype is the result of the interaction of genetic and environmental factors during its lifetime (including during development). The expression of genes in an organism can be influenced by both the internal and external environment both during and after development.

Dominant, recessive, codominant, and multiple alleles, DNA modifications (such as methylation), and interactions between genes, combine in their effects.

Don Horne

Original DNA

Genotype

Determines the genetic potential of an individual

Environmental factors

Variation is the essential raw material for natural selection. Different phenotypic variants will have different relative survival and reproductive success (fitness) in the prevailing environment and only the most successful variants will proliferate.

The external and internal environments can influence the expression of the genotype. The external environment includes physical factors such as temperature, or biotic factors such as competition. The internal environment, e.g. presence or absence of hormones or growth factors during development, may also affect genotypic expression.

(b) How does the environment contribute to the phenotype we see:

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1. (a) What is the basis of the genetic variation of sexually reproducing organisms?

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Quantitative traits are characterised by continuous variation, with individuals falling somewhere on a normal distribution curve of the phenotypic range. Typical examples include skin colour and height in humans (left), grain yield in corn (above), growth in pigs (above, left), and milk production in cattle (far left). Quantitative traits are determined by genes at many loci (polygenic) but most are also influenced by environmental factors.

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Albinism (above) is the result of the inheritance of recessive alleles for melanin production. Those with the albino phenotype lack melanin pigment in the eyes, skin, and hair. Comb shape in poultry is a qualitative trait (fits into a category) and birds have one of four phenotypes depending on which combination of four alleles they inherit. The dash (missing allele) indicates that the allele may be recessive or dominant.

Single comb rrpp

Walnut comb R_P_

Pea comb rrP_

Rose comb R_pp

Flower colour in snapdragons (right) is also a qualitative trait determined by two alleles (for red and white). The alleles show incomplete dominance and the heterozygote (CRCW) exhibits an intermediate phenotype between the two homozygotes. CRCR

CWCW

2. Why is phenotypic variation important in the process of natural selection?

3. (a) What is a neutral mutation?

(b) What is the significance of neutral mutations?

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4. Describe the differences between continuous and discontinuous variation, giving examples to illustrate your answer:

5. Identify each of the following phenotypic characteristics as continuous (quantitative) or discontinuous (qualitative): (a) Wool production in sheep: (d) Albinism in mammals: (b) Hand span in humans:

(e) Body weight in mice:

(c) Blood groups in humans:

(f) Flower colour in snapdragons:

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97 Processes in Gene Pools

Key Idea: The proportions of alleles in a gene pool can be altered by the processes that increase or decrease variation. This activity portrays two populations of a beetle species. Each beetle is a 'carrier' of genetic information, represented by the alleles (A and a) for a gene that controls colour and has a dominant/recessive inheritance pattern. There are normally

two phenotypes: black and pale. Mutations may create new alleles. Some of the microevolutionary processes (natural selection, genetic drift, gene flow, and mutation) that affect the genetic composition (allele frequencies) of gene pools are shown below. Simulate the effect of these processes using the cut out, Gene Pool Exercise. Mutations: Mutations can create new alleles. Mutation is the original source of genetic variation that provides new material for natural selection.

An aspect of gene flow. Populations can gain alleles from other gene pools.

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Immigration

Aa

AA

Aa

E

aa

m

on a ti igr

An aspect of gene flow. Genes may be lost to other gene pools.

A'A

AA

Aa

AA

Aa

Aa

AA

aa

Aa

Aa

aa

AA

Aa

aa

aa

Aa

AA

Deme 1

AA

Aa

A deme is a local population that is genetically isolated from other local populations of the species.

Natural selection: Selection against unfavourable allele combinations may reduce survival or reproductive success. Natural selection accumulates and maintains favourable genotypes, reduces genetic diversity within the gene pool, and increases differences between populations.

Geographical barriers isolate the gene pool and prevent regular gene flow between populations.

Key to genotypes and phenotypes

Gene flow: Genes are exchanged with other gene pools as individuals move between them. Gene flow is a source of new genetic variation and tends to reduce differences between populations that have accumulated because of natural selection or genetic drift.

Aa

AA

Black

Homozygous dominant

Aa

Black

Heterozygous

aa

A'A

Pale

Mottled

Homozygous Homozygous recessive dominant (mutant)

Deme 2

Aa

aa

Aa

aa

Aa

Aa

Aa

Aa

Aa

aa

Aa

aa

aa

aa

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aa

AA

Aa

Aa

Boundary of gene pool

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Genetic drift: Chance events cause the allele frequencies of small populations to ‘drift’ (change) randomly from generation to generation. Genetic drift has a relatively greater effect on the genetics of small populations and can be important in their evolution. Small populations may occur as a result of the founder effect (where a small number of individuals colonise a new area) or genetic bottlenecks (where the population size is dramatically reduced by a catastrophic event) . LINK

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Mate choice: Individuals may not select their mate randomly and may seek out particular phenotypes, increasing the frequency of the associated alleles in the population.

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1. For each of the two demes shown on the previous page (treating the mutant in deme 1 as a AA):

(a) Count up the numbers of allele types (A and a).

(b) Count up the numbers of allele combinations (AA, Aa, aa).

Allele types

Number counted

Deme 2

%

A

A

Allele types

a

a

AA Allele combinations

%

AA Allele combinations

Aa

Aa

aa

aa

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2. Calculate the frequencies as percentages (%) for the allele types and combinations:

Number counted

Deme 1

3. One of the fundamental concepts for population genetics is that of genetic equilibrium, stated as: "For a very large, randomly mating population, the proportion of dominant to recessive alleles remains constant from one generation to the next". If a gene pool is to remain unchanged, it must satisfy all of the criteria below that favour gene pool stability. Few populations meet all (or any) of these criteria and their genetic makeup must therefore be continually changing. For each of the five factors (a-e) below, state briefly how and why each would affect the allele frequency in a gene pool:

(a) Population size:

Factors favouring gene pool stability

Aa

Aa aa AA

Aa

(b) Mate selection:

Aa

AA

AA

Aa

Aa

Aa

Aa

AA

Aa

AA

Aa

aa

Aa Aa

aa

aa

Aa

Aa aa

Aa

aa

Aa

Aa

Aa

AA

Factors favouring gene pool change

Aa Aa

Aa

AA

Aa

Aa

Aa

aa

Aa Aa

aa

Aa

Aa

(c) Gene flow between populations:

AA

aa

Aa

aa

aa

aa

AA

Aa

Aa AA

Barrier to gene flow

Aa AA Aa

Aa

Aa

aa

aa

aa

AA

AA

Aa

Aa

Assortative mating

Aa

AA

Aa

Aa

aa

Aa

Aa aa

AA

Aa

AA

Aa

aa

4. Identify the factors that tend to:

aa

Aa

Aa

Aa

AA

Emigration

Aa

New recessive allele

Aa

AA

Aa

Aa Aa

Aa aa

aa Aa

AA aa

Aa AA

aa

Aa

AA

AA Aa

AA

Aa

AA

aa aa

aa

Aa

aa

Aa Aa AA

AA

Aa

Aa

AA

Mutations

Aa

AA Aa

Aa

a'a

No mutation

aa

Aa

Aa AA aa AA AA Aa aa Aa aa AA AA Aa aa Aa Aa

AA

AA

aa

AA

Gene flow

Aa

AA

Aa

Aa

Aa

Aa

(a) Increase genetic variation in populations:

(b) Decrease genetic variation in populations:

Aa

aa Aa

AA

Immigration

Aa AA

No gene flow

(e) Natural selection:

Aa

aa AA

Aa

AA

Aa

Aa AA

aa

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aa

Aa

aa AA AA

aa

Aa

AA

No natural selection

aa

AA

Aa

Aa

AA

Aa Aa

Aa

AA

Aa

Aa

aa

Aa

AA

Aa

AA

Aa

Aa

aa

Aa AA

aa

Aa

AA

Natural selection

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(d) Mutations:

AA

AA

Aa

aa

Aa

Random mating

Aa

AA

AA

aa

Aa

Small population

AA

AA

aa

Aa AA

Aa

Large population

aa

Aa AA

aa

aa Aa

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98 What is a Gene Mutation?

Key Idea: Gene mutations are localised changes to the DNA base sequence. Gene mutations are small, localised changes in the DNA base sequence caused by a mutagen or an error during DNA replication. The changes may involve a single nucleotide (a point mutation) or a triplet. Point mutations can occur by substitution, insertion, or deletion of bases and alter the

No mutation G

C

T

C

T

T

C

G

A

G

A

A

PR E O V N IE LY W

Silent mutations

mRNA transcribed. A point mutation may not alter the amino acid sequence because the degeneracy of the genetic code means that more than one codon can code for the same amino acid. Such mutations are called silent because the change is not recorded in the amino acid sequence. Mutations that do result in a change in the amino acid sequence will most often be harmful because they alter protein functionality.

Silent mutations do not change the amino acid sequence nor the final protein. In the genetic code, several codons may code for the same amino acid (this so-called degeneracy creates redundancy in the genetic code).

Silent mutations are also neutral if they do not alter fitness, although it is now known that so-called silent changes may still affect mRNA stability and transcription, even though they do not change codon information. In these cases, the changes will not be neutral.

Original DNA

Transcription

Arg

Glu

Normal sequence

mRNA

Translation

G

C

C

C

T

T

C

G

G

G

A

A

Amino acids

Phe

Glu

Tyr

Glu

Val

Leu

Amino acid sequence forms a normal polypeptide chain

Arg

Glu

Mutated sequence

Missense substitution

A single base is substituted for another base. Some substitutions may still code for the same amino acid (because of redundancy in the genetic code), but they may also result in a codon that codes for a different amino acid. In the example (centre right), placing a T where a C should have been in the normal sequence (top right), results in the amino acid lysine appearing where glutamic acid should be. This could affect this protein’s function. If a missense substitution results in an amino acid with similar chemical properties, or the change affects a non-critical part of the protein, the change may be effectively neutral.

Mutation: Substitute T instead of C

Mutant DNA mRNA

Amino acids

Phe

Lys

Tyr

Glu

Val

Leu

Polypeptide chain with wrong amino acid

Mutation: Substitute A instead of C

Nonsense substitution

Some codon changes are more likely than others to cause a change in the amino acid encoded by the sequence (amino acids encoded by 4 or 6 different codons are less likely to be affected by substitutions). In the example illustrated, a single base substitution in the first nucleotide of the third codon has a dramatic effect on the nature of the encoded polypeptide chain. The codon no longer codes for an amino acid, but instead terminates the translation process of protein synthesis. This results in a very short polypeptide chain that is likely to have little or no function because the STOP codon is introduced near the START codon.

Mutant DNA mRNA

Amino acids

Phe

Tyr

Mutated DNA creates a STOP codon which prematurely ends synthesis of the polypeptide chain

1. Why are mutations that alter the amino acid sequence usually harmful?

(b) When is a silent mutation considered to be neutral?

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3. (a) What is a silent mutation?

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2. Why are nonsense substitutions more damaging than missense substitutions?

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99 Reading Frame Shifts

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146

Key Idea: Point mutations may cause a reading frame shift. These usually produce a non-functional protein. A frame shift mutation occurs when bases are inserted or deleted from a DNA sequence and the total nucleotide sequence is no longer a multiple of three. All of the nucleotides

Normal sequence

Original DNA

PR E O V N IE LY W

All examples of mutations below are given with reference to the normal sequence and its encoded amino acid sequence (right).

after the frame shift mutation are displaced, creating a new reading frame sequence of codons. In most cases, there will be a change in the amino acids produced, and the protein produced will be non-functional or incorrect. Many diseases are the result of frame shift mutations.

mRNA

Amino acids

Phe

Tyr

Glu

Glu

Val

Leu

Amino acid sequence forms a normal polypeptide chain Insertion of C

Reading frame shift by insertion

The insertion of a single extra base into a DNA sequence displaces the bases after the insertion by one position. In the example (right) instead of the next triplet being CTT it is CCT, and a different amino acid is produced.

Mutant DNA

NOTE: could also lead to nonsense

Amino acids

mRNA

Phe

Tyr

Gly

Arg

Gly

Ser

Reading frame shift results in a new sequence of amino acids. The protein is unlikely to have any biological activity. Deletion of C

Reading frame shift by deletion

The deletion of an nucleotide also causes a frame shift. Again the result is usually a polypeptide chain of doubtful biological activity.

Mutant DNA

NOTE: could also lead to nonsense

mRNA

Mutations that cause a shift in the reading frame almost always lead to a non-functional protein, but the effect is reduced the further the mutation is from the START codon.

Amino acids

Tyr

Lys

Arg

Phe

?

Reading frame shift results in a new sequence of amino acids. The protein is unlikely to have any biological activity. Insertion of C

Partial reading frame shift

Both an insertion and a deletion of bases within a gene can cause a frame shift effect in which each codon no longer has the correct triplet of three bases. In this example, three codons have been affected, along with the amino acids they code for. The error is limited to the codons between the insertion and deletion. There is no biological activity if the amino acids altered are important to the functioning of the resulting protein.

Phe

Deletion of C

Mutant DNA mRNA

Amino acids

Phe

Val

Arg

Lys

Val

Leu

Altered chain which may or may not produce a protein with biological activity.

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1. What is a reading frame shift?

2. Why is a frame shift near the start codon likely to have a greater impact than one near the stop codon?

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3. What is the effect of a partial reading shift mutation on the final protein produced:

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100 Chromosome Mutations

Key Idea: Large scale mutations occurring during meiosis can fundamentally change chromosome structure. Chromosome mutations (also called block mutations) involve the rearrangement of whole blocks of genes (involving many bases), rather than individual bases within a gene. They commonly occur during meiosis and they alter the number

or sequence of whole sets of genes on the chromosome (represented by letters below). Translocations may sometimes involve the fusion of whole chromosomes, thereby reducing the chromosome number of an organism. This is thought to be an important evolutionary mechanism by which instant speciation can occur.

Deletion

Inversion

Break

Break

Genes

Break

Genes

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Break

Step 1

A B C D E F G H

Step 1

M N O P Q R S T

A B C D E F G H

Segment rotates 180 °

C

Segment is lost

C D E F

M N O P Q R S T

D

G H

Step 2

M N O P Q R S T

A B

A B

G H

M N O P Q R S T

A B F E D C G H

Step 3

A break may occur at two points on the chromosome and the middle piece of the chromosome falls out. The two ends then rejoin to form a chromosome deficient in some genes. Alternatively, the end of a chromosome may break off and is lost.

Step 1

5 6 7 8 9 0

Step 1

M N O P Q R S T

Fragment joins on to homologous chromosome

Break

A B C D E F

Step 2

5 6 7 8 9 0

1 2 3 4 G H

Step 2

M N O P Q R S T

G H

M N O P Q

A B C D E F

M N O P Q

A B C D E F

M N O P Q

F

M N O P Q

Segment removed

Segments join

A B C D E F 1 2 3 4

A B C D E F

Break

A B C D E

Segment removed

Step 3

M N O P Q R S T

Duplication

1 2 3 4

A B C D E F G H

M N O P Q R S T

The middle piece of the chromosome falls out and rotates through 180° and then rejoins. There is no loss of genetic material. The genes will be in a reverse order for this segment of the chromosome.

Translocation

Non-homologous chromosomes pair during meiosis

G H

Segment rejoins

Chromosome rejoins

Step 3

F

A B

E

Step 2

5 6 7 8 9 0

Step 3

A B C D E A B C D E F

M N O P Q

F

M N O P Q

M N O P Q R S T

In translocation mutations, a group of genes moves between different chromosomes. The large chromosome (white) and the small chromosome (blue) are not homologous. A piece of one chromosome breaks off and joins to the other. When the chromosomes are passed to gametes, some gametes will receive extra genes, while some will be deficient.

A segment is lost from one chromosome and is added to its homologue. In this diagram, the darker chromosome on the bottom is the 'donor' of the duplicated piece of chromosome. The chromosome with the segment removed is deficient in genes. Some gametes will receive double the genes while others will have no genes for the affected segment.

1. Which of the chromosome mutations above results in a loss of genetic information?

2. For each of the chromosome (block) mutations below, write the new gene sequence after the mutation has occurred: Original sequence

Mutated sequence

A B C D E F G H M N O P Q R S T

1

(b) Translocation:

2

3

4

5

6

7

8

9

0

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(a) Inversion:

4. Why do translocations sometimes reduce the total number of chromosomes?

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3. Which type of block mutation is likely to be the least damaging to the organism? Explain your answer:

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101 Gene Duplication as a Source of New Genes The duplication of a gene frees one of those genes to develop a new function, while the original continues as it was. In the case of genes that produce proteins with a tendency or ability to perform two functions, there may be adaptive conflict, in which its ability to perform one function compromises its ability to perform another. Gene duplication solves this problem by allowing natural selection to act on the genes so that they follow different evolutionary paths.

Caenorhabditis elegans (nematode): 49% genes duplicated

Saccharomyces cerevisiae (yeast): 30% genes duplicated

Drosophila melanogaster (fruit fly): 41% genes duplicated

CDC

Haemophilus influenza (bacterium): 17% genes duplicated

Homo sapiens: 38% genes duplicated Total gene number

Number of duplicated genes (%)

Bacteria

Mycoplasma pneumoniae Helicobacter pylori

677

298 (44)

1590

266 (17)

Archaea

Archaeoglobus fulgidus

2436

719 (30)

Eukarya

Arabidopsis thaliana (thale cress)

25,498

16,574 (65)

Gene duplication in colobine monkeys has enabled the production of enzymes that optimally perform similar functions in different body environments. The primary food source of colobines, unlike most other primates, is leaves. The leaves are fermented in the gut by bacteria, which are then digested with the assistance of an enzyme produced by RNase genes. In colobines there are two forms of the RNase genes, RNase1 and RNase1B, while in other primates there is only RNase1. The optimal pH for the enzyme RNase1 and RNase1B are 7.4 and 6.3 respectively. In colobine monkeys, the pH of the digestive system is 6-7, but in other primates it is 7.4-8. RNase1B is six times more efficient at degrading RNA in the gut of colobines than RNase1. RNase1 is also expressed in cells outside the digestive system where it degrades double stranded RNA and may assist in defense against viral infection. RNase1B is 300 times less efficient at this function.

Olivier Lejadei CC 2.0

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Key Idea: Gene duplication can provide new genes for natural selection to act on. Gene duplication can potentially create more genetic information, which can then be influenced by selection pressures. Studies have shown that gene duplication has occurred in virtually every species and often more than once. It is well known that many crop plants (e.g. kiwifruit) have duplicated not just one gene, but their entire genome.

Many fish living in the near AFP III freezing waters of the Antarctic ensure their blood remains ice free by employing proteins with antifreeze properties. There are four major antifreeze proteins used by fish (labeled AFP types I - IV). The gene for the protein AFP III, found in Antarctic eelpout, is very similar to the gene that produces sialic acid synthase (SAS) . Molecular studies have found that a slight modification in the SAS gene causes the production and secretion of AFP III. More importantly, the SAS gene also shows ice binding capabilities. It appears that a duplication of the SAS gene produced a new gene that was selected for its ice binding capabilities and thus diverged to become the AFP III gene in Antarctic eelpout.

Adapted from J. Zhang, University Michigan 2003

(b) Explain how duplication can resolve adaptive conflict:

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1. (a) Use an example to explain how gene duplication allows for the evolution of new gene function:

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102 Aneuploidy gamete receives two of the same type of chromosome and the other gamete receives no copy. This non-disjunction results in abnormal numbers of chromosomes passing to the gametes. Aneuploids form if there is non-disjunction in one chromosome. For example, Down syndrome is an aneuploid caused by non-disjunction of chromosome 21 and individuals have three copies of that chromosome. This is a type of polysomy called trisomy and is described as 2N +1.

Non-disjunction in meiosis I

Non-disjunction in meiosis II

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Key Idea: Aneuploidy is the condition where the total chromosome number is not an exact multiple of the normal haploid set for the species. The meiotic spindle normally distributes chromosomes to daughter cells without error. However, mistakes can occur in which the homologous chromosomes fail to separate properly at anaphase during meiosis I, or sister chromatids fail to separate during meiosis II. In these cases, one

Meiosis I

24

N+1

Meiosis II

24

23

N

23

23

N

23

22

N-1

24

N+1

Meiosis I

24

N+1

46

Meiosis II

46

22

Non-disjunctions show a maternal age effect (frequency increases with maternal age).

22

N-1

22

N-1

Number of chromosomes

Gametes

Gametes

Agnieszka Kwiecie, license: CC-BY 3.0

Edward syndrome

Patau syndrome

Datura stramonium

Edward syndrome has an incidence rate of 1 in 5000 live births (with a maternal age effect). Features include severe mental retardation, low set, malformed ears, congenital heart defects, small mouth and rocker-bottom feet. Half the patients die by two months of age, and only a few have been known to survive beyond several years of age. About 95% of the affected foetuses are spontaneously aborted.

Patau syndrome has an incidence rate of 1 in 3000 live births (with a maternal age effect). All patients have markedly retarded mental and physical development, including various eye defects, cleft palate and cleft lip, polydactyly, low set, malformed ears and a variety of defects of internal organs. About half the affected babies die by one month of age with a rare instance of one patient surviving to age 10 years.

The plant Datura stramonium has 12 sets of chromosomes. There are 12 known aneuploids, each trisomic for a different chromosome. Interestingly each aneuploid has its own variety of seed pod shape, ranging from buckling (trisomy 3) to cocklebur (trisomy 6) and spinach (trisomy 10). All the aneuploids survive to be viable adult plants indicating plants are better able to accommodate genetic shuffling.

2. (a) Edward syndrome is caused by a trisomy of which chromosome?

(b) Patau syndrome is caused by a trisomy of which chromosome?

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1. Which affects the greater number of gametes, non-disjunction at meiosis I or non-disjunction at meiosis II?

3. What is the chromosome number for each of the human examples pictured above?

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4. Contrast the effects on viability of trisomy in a plant such as Datura and a human?

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103 Polyploidy

Key Idea: Polyploidy is a condition in which a cell or organism contains three or more times the haploid number of chromosomes (3N or more). Polyploidy is rare in animals, but more common in plants. Natural polyploid species occur in animals particularly among hermaphrodites (those having both male and female sex organs) such as flatworms and earthworms. Animals with

parthenogenetic females (which produce viable offspring without fertilisation) can be polyploid (e.g. some beetles, moths, shrimp, goldfish, and salamanders). Polyploid species occur in all major groups of plants. They are common in ferns and occur in about 47% of all flowering plants. There are two types of polyploidy recognised: allopolyploidy and autopolyploidy.

Autopolyploidy

PR E O V N IE LY W

Allopolyploidy

Species A

AA

A

Species B

Infertile hybrid

BB

AB

B

Same species

AB

Non-disjunction doubles the chromosome number in the hybrid.

AB

AABB

Species C

AA

AA

AA

A

AA

AA

Haploid gametes

Infertile hybrid reproduces asexually.

AABB

Same species

Normal haploid gamete

This type of polyploidy results from mating between two species. The resulting hybrid, with chromosomes from each of the parent species, may be sterile. Non-disjunction in the sterile hybrid can result in all chromosomes having a homologue with which to pair during meiosis. Self-fertilisation may then produce a viable, fertile hybrid. Allopolyploids show greater genetic variability and vigour than autopolyploids.

Many commercial plant varieties are allopolyploids. Their increased hybrid vigour relative to the parental genotypes is the result of many factors, including environmental influences on the genotype (epigenetics), and masking of deleterious alleles. Polyploidy also results in gene redundancy and so provides the ability to diversify gene function so that extra copies of genes might end up being used in entirely different ways.

Common wheat 6N = 42

Diploid gametes

AA

Diploid gamete

AAA

Union of gametes from this hybrid produces a new species of interbreeding plants: a fertile allopolyploid.

AA

Sterile hybrid

AAAA

Fertile hybrid

Autopolyploidy refers to the multiplication of one basic set of chromosomes. It occurs when chromosomes fail to separate during meiosis or from the failure of the cell to divide after the chromatids have separated. The karyotype possesses chromosomes only from one species. Two forms are shown above, one sterile, the other fertile. The total chromosome complement is represented by a multiple of identical sets.

Tobacco 4N = 48

Banana 3N = 27

Boysenberry 7N = 49

Strawberry 8N = 56

2. (a) Identify which of the polyploid plants shown above are sterile:

(b) What effect might this have on the genetic variation of the population?

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1. Distinguish between an allopolyploid and an autopolyploid:

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3. Explain why polyploids might have a selective advantage over the parental genotypes:

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104 Mutation in Populations: Sickle Cell

Key Idea: The substitution of one nucleotide from T to A results in sickle cell disease. The mutation is codominant. Sickle cell disease is an inherited blood disorder caused by a gene mutation (Hbs), which produces a faulty beta (β) chain haemoglobin (Hb) protein. This in turn produces red blood cells Normal haemoglobin produces normal red blood cells

Mutant haemoglobin produces sickle-shaped red blood cells

Sickle cells

PR E O V N IE LY W

Normal red blood cells

with a deformed sickle appearance and a reduced capacity to carry oxygen. Many aspects of metabolism are also affected. The mutation is codominant (both alleles equally expressed), and people heterozygous for the mutation (carriers) have enough functional haemoglobin and suffer only minor effects.

Each red blood cell (RBC) contains about 270 million haemoglobin molecules. In their normal state, the red blood cells have a flattened disc shape which allows them to squeeze through capillaries to offload their oxygen to tissues.

The HBB Gene The gene coding for the β-chain of haemoglobin is on chromosome 11 and consists of 438 bases.

p

HBB gene

Each haemoglobin molecule is made up of two α-chains and two β-chains linked together

The 438 nucleotides produce a protein made up of 146 amino acids

The mutated form of haemoglobin has reduced solubility and precipitates when deprived of oxygen. This deforms the red blood cells giving them a rigid sickle shape, which prevents their movement through capillaries.

Sickle cell anaemia

The sickled RBCs are removed from the circulation leading to anaemia. Their rigid shape blocks small vessels and leads to widespread tissue and organ damage.

β-chain haemoglobin

First base

Normal base: T Substituted base: A

q

DNA

Code corresponding to the 1st amino acid

This sequence is the beginning of the DNA template strand for a normal β-chain of haemoglobin (excluding start sequence TAC). The sickle cell mutation involves the substitution of one base for another in the HBB gene, causing one amino acid to be altered. This new amino acid is hydrophobic rather than hydrophilic, which makes the Hb collapse in on itself when deprived of oxygen.

Sickle cell and malaria

The sickle cell mutation (HbS) is lethal in the homozygote (two mutated alleles present) but heterozygotes (one mutated allele) are much less susceptible to malaria than unaffected people (two normal alleles). This is because the malarial parasite cannot infect the deformed blood cells. A high frequency of the mutation is present in many regions where malaria is endemic (present in the population all the time). The ability of a heterozygote to confer an adaptive advantage like this is called heterozygous advantage.

(b) How does the sickle cell mutation result in the symptoms of the disease?

(c) Explain why heterozygotes (carriers) suffer only minor effects:

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1. (a) Explain the genetic cause of sickle cell disease:

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2. Briefly explain why there is a high frequency of the sickle cell mutation in populations where malaria is endemic:

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105 Mutation in Populations: Cystic Fibrosis

Key Idea: Cystic fibrosis most often results from a triplet deletion in the CFTR gene, producing a protein that is unable to regulate chloride transport. Cystic fibrosis (CF) is an inherited disorder caused by a mutation of the CFTR gene. It is one of the most common lethal autosomal recessive conditions affecting people of European descent (4% are carriers). The CFTR gene's protein

product is a membrane-based protein that regulates chloride transport in cells. Over 500 mutations of the CFTR gene are known, causing disease symptoms of varying severity. The D(delta)F508 mutation accounts for more than 70% of all defective CFTR genes. This mutation leads to an abnormal CFTR, which cannot take its proper position in the membrane (below) nor perform its transport function.

Abnormal CFTR (1479 amino acids)

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Normal CFTR (1480 amino acids)

Correctly controls chloride ion balance in the cell

No or little control of chloride ion balance in the cell Mucus build up

Cell exterior Cell exterior

Chloride ions effectively removed from the cell.

CFTR protein

Normal CFTR (1480 amino acids)

Abnormal CFTR (1479 amino acids)

Correctly controls chloride ion balance in the cell

Unable to control chloride ion balance in the cell

Cl-

Cl-

Cellinterior interior Cell

Water Chloride ions are effectively removed from the cell.

Cl-

Normal CFTR

Cl-

-

Cl

Water

Water

Abnormal CFTR

Cl-

Outside the cell

The DF508 mutation causes the Plasma membranerapidly, CFTR protein to degrade stopping itCell from inserting into the cytoplasm plasma membrane.

Chloride channel

Chloride channel

Chloride ions build up inside the cell.

-

Chloride ions build Cl- Cl Clup inside the cell. Cl Cl

Cl-

p

q

Cl-

Cl

-

The CF gene on chromosome 7

CFTR protein

The CF gene is located on chromosome 7. The DF508 mutation of the CF gene describes a deletion of the 508th triplet, which in turn causes the loss of a single amino acid from the gene’s protein product, the cystic fibrosis transmembrane conductance regulator (CFTR). This protein normally regulates the chloride channels in cell membranes, but the mutant form fails to achieve this. The portion of the DNA containing the mutation site is shown below:

The DF508 mutant form of CFTR fails to take up its position in the membrane. Its absence results in defective chloride transport and leads to a net increase in water absorption by the cell. This accounts for the symptoms of cystic fibrosis, where mucus-secreting glands, particularly in the lungs and pancreas, become fibrous and produce abnormally thick mucus. The widespread presence of CFTR throughout the body also explains why CF is a multisystem condition affecting many organs.

The CFTR protein consists of 1480 amino acids

Base 1630

CFTR gene

DNA template strand

This triplet codes for the 500th amino acid

These three bases are deleted from the 507/508th triplets in the ΔF508 mutation

1. (a) Write the mRNA sequence for the transcribing DNA strand above:

(c) What kind of mutation is DF508?

2. (a) Explain why the abnormal CFTR fails to transport Cl- correctly:

(b) What effect does this have on water movement in and out of the cell?

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(b) Rewrite the mRNA sequence for the mutant DNA strand:

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106 Beneficial Mutations in Humans population. This is because the mutations have been in existence for a relatively short time, so the mutations have not had time to become widespread in the human population. Scientists often study mutations that cause disease. By understanding the genetic origin of various diseases it may be possible to develop targeted medical drugs and therapies against them.

The village of Limone, Italy

Structurally ApoA1 is composed of eight a-helices that form a twisted loop

Apolipoprotein A1-Milano is a well documented mutation to apolipoprotein A1 that helps transport cholesterol through the blood. The mutation causes a change to one amino acid from arginine at position 173 to cysteine. This single change increases the protein's effectiveness by ten times, dramatically reducing incidence of heart disease. The mutation can be traced back to its origin in Limone, Italy, in 1644. Another mutation to a gene called PCSK9 has a similar effect, lowering the risk of heart disease by 88%.

Red blood cell containing haemoglobin

Malarial parasite

Lactose is a sugar found in milk. All infant mammals produce an enzyme called lactase that breaks the lactose into the smaller sugars glucose and galactose. As mammals become older, their production of lactase declines and they lose the ability to digest lactose. As adults, they become lactose intolerant and feel bloated after drinking milk. About 10,000 years ago a mutation appeared in humans that maintained lactase production into adulthood. This mutation is now carried in people of mainly European, African and Indian descent.

Malaria resistance results from a mutation to the haemoglobin gene (HbS) that also causes sickle cell disease. This mutation has some benefit in regions where malaria is common. A less well known mutation (HbC) to the same gene, discovered in populations in Burkina Faso, Africa, results in a 29% reduction in the likelihood of contracting malaria if the person has one copy of the mutated gene and a 93% reduction if the person has two copies. In addition, the anaemia that person suffers as a result of the mutation is much less pronounced than in the HbS mutation.

1. Why is it that many of the recent beneficial mutations recorded in humans have not spread throughout the entire human population?

3. Why would it be beneficial to be able to digest milk in adulthood?

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2. What selection pressure could act on apolipoprotein A1-Milano to help it spread through a population?

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Key Idea: Beneficial mutations increase the fitness of the organisms that possess them but are relatively rare. Beneficial mutations are mutations that increase the fitness of the organisms that possess them. Although beneficial mutations are rare compared to those that are harmful, there are a number of well documented beneficial mutations in humans. Some of these mutations are not very common in the human

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107 Mechanism of Natural Selection

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Key Idea: Natural selection is the evolutionary mechanism by which organisms that are better adapted to their environment survive to produce a greater number of offspring. Evolution is the change in inherited characteristics in a population over generations. Evolution is the consequence

Natural selection is the term for the

Darwin's theory of evolution by natural selection Darwin's theory of evolution by natural selection is outlined below. It is widely accepted by the scientific community today and is one of founding principles of modern science.

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mechanism by which better adapted organisms survive to produce a greater number of viable offspring. This has the effect of increasing their proportion in the population so that they become more common. This is the basis of Darwin's theory of evolution by natural selection.

of interaction between four factors: (1) The potential for populations to increase in numbers, (2) Genetic variation as a result of mutation and sexual reproduction, (3) competition for resources, and (4) proliferation of individuals with better survival and reproduction.

We can demonstrate the basic principles of evolution using the analogy of a 'population' of M&M's candy.

#1

In a bag of M&M's, there are many colours, which represents the variation in a population. As you and a friend eat through the bag of candy, you both leave the blue ones, which you both dislike, and return them to bag.

Overproduction

Variation

Populations produce too many young: many must die

Individuals show variation: some variations more favourable than others

Populations generally produce more offspring than are needed to replace the parents. Natural populations normally maintain constant numbers. A certain number will die without reproducing.

Individuals in a population have different phenotypes and therefore, genotypes. Some traits are better suited to the environment, and individuals with these have better survival and reproductive success.

Natural selection

Natural selection favours the individuals best suited to the environment at the time

#2

The blue candy becomes more common...

Individuals in the population compete for limited resources. Those with favourable variations will be more likely to survive. Relatively more of those without favourable variations will die.

Inherited

Variations are inherited: the best suited variants leave more offspring

#3

1. Identify the four factors that interact to bring about evolution in populations:

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The variations (both favourable and unfavourable) are passed on to offspring. Each generation will contain proportionally more descendants of individuals with favourable characters.

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Eventually, you are left with a bag of blue M&M's. Your selective preference for the other colours changed the make-up of the M&M's population. This is the basic principle of selection that drives evolution in natural populations.

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1. Variation through mutation and sexual reproduction: In a population of brown beetles, mutations independently produce red colouration and 2 spot marking on the wings. The individuals in the population compete for limited resources.

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Variation, selection, and population change

Red

Brown mottled

Red 2 spot

2. Selective predation: Brown mottled beetles are eaten by birds but red ones are avoided.

3. Change in the genetics of the population: Red beetles have better survival and fitness and become more numerous with each generation. Brown beetles have poor fitness and become rare.

Natural populations, like the ladybug population above, show genetic variation. This is a result of mutation (which creates new alleles) and sexual reproduction (which produces new combinations of alleles). Some variants are more suited to the environment of the time than others. These variants will leave more offspring, as described for the hypothetical population (right).

2. What produces the genetic variation in populations? 3. Define evolution:

4. Explain how the genetic make-up of a population can change over time:

1 2 3

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Beetle population

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5. Complete the table below by calculating the percentage of beetles in the example above right.


108 Adaptation and Fitness

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Key Idea: Adaptive features enhance an individual's fitness. An adaptation is any heritable trait that suits an organism to its natural role in the environment (its niche). These traits may be structural, physiological, or behavioural. Adaptations promote fitness, which is mathematical measure of the contribution an organism makes to the next generation. Adaptations are distinct from properties. Although they may

Number of horns in rhinoceroses

be striking, properties cannot be described as adaptive unless they are shown to be functional in the organism’s natural habitat. Genetic adaptation must not be confused with acclimatisation, which refers to an individual's ability to alter its behaviour or physiology in response to its changing environment (e.g. adjustment to altitude in mountaineers). Examples of adaptive features are illustrated below.

Great Indian rhino

Krish Dulal CC 3.0

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African black rhino

Harald Zimmer CC3.0

Not all differences between species can be convincingly interpreted as adaptations to particular environments. Rhinoceroses charge rival males and predators, and the horn(s), when combined with the head-down posture, add effectiveness to this behaviour. Horns are obviously adaptive, but it is not clear that the possession of one (Indian rhino) or two (black rhino) horns is necessarily related directly to the environment in which those animals live.

Ear length in rabbits and hares

The external ears of many mammals are used as important organs to assist in thermoregulation (controlling loss and gain of body heat). The ears of rabbits and hares native to hot, dry climates, such as the jack rabbit of south-western USA and northern Mexico, are relatively very large. The Arctic hare lives in the tundra zone of Alaska, northern Canada and Greenland, and has ears that are relatively short. This reduction in the size of the extremities (ears, limbs, and noses) is typical of cold adapted species.

Black-tail jackrabbit: Lepus californicus

Arctic hare: Lepus arcticus

Adaptive features

Phalanger

Adaptation: Flap of skin extends from hind legs to fore legs, increasing the surface area of the body and assisting gliding.

Adaptation: Tail is used as a rudder in swimming, but its main function is to serve as a fat storage area.

Platypus

Adaptation: Webbed forefeet are the main propulsive organs in swimming (the webs can be folded back when walking or burrowing).

Marsupial mole

Adaptation: The soft and leathery bill, which is sensitive to touch, electrical activity, movements and vibrations, is used for navigation and for finding food.

Adaptation: The claws of the fore legs are highly modified for digging into soil and thrusting the loosened dirt backwards.

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2. Explain the evolutionary significance of adaptive features:

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1. Distinguish between adaptive features (genetic) and acclimatisation:

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109 Gene Pool Exercise

Cut out each of the beetles below and use them to reenact different events within a gene pool as described in this topic

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pages: Gene Pools and Evolution, Changes in a Gene Pool, The Founder Effect, Population Bottlenecks, Genetic Drift.

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110 Changes in a Gene Pool

Key Idea: Natural selection and migration can alter the allele frequencies in gene pools. The diagram below shows an hypothetical population of beetles undergoing changes as it is subjected to two ‘events’. The three phases represent a progression in time (i.e. the

same gene pool, undergoing change). The beetles have two phenotypes (black and pale) determined by the amount of pigment deposited in the cuticle. The gene controlling this character is represented by two alleles A and a. Your task is to analyse the gene pool as it undergoes changes.

1. For each phase in the gene pool below fill in the tables provided as follows; (some have been done for you): (a) Count the number of A and a alleles separately. Enter the count into the top row of the table (left hand columns). (b) Count the number of each type of allele combination (AA, Aa and aa) in the gene pool. Enter the count into the top row of the table (right hand columns). (c) For each of the above, work out the frequencies as percentages (bottom row of table):

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Allele frequency =

No. counted alleles ÷ Total no. of alleles x 100

Phase 1: Initial gene pool

Black

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Allele types Allele combinations

Two pale individuals died. Their alleles are removed from the gene pool.

Phase 2: Natural selection

In the same gene pool at a later time there was a change in the allele frequencies. This was due to the loss of certain allele combinations due to natural selection. Some of those with a genotype of aa were eliminated (poor fitness).

These individuals (surrounded by small white arrows) are not counted for allele frequencies; they are dead!

A

a

AA

Aa

aa

No. No. %%

This individual is entering the population and will add its alleles to the gene pool.

Phase 3: Immigration and emigration

This individual is leaving the population, removing its alleles from the gene pool.

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No. No. %%

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Individuals coming into the gene pool (AA) are counted for allele frequencies, but those leaving (aa) are not.

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This particular kind of beetle exhibits wandering behaviour. The allele frequencies change again due to the introduction and departure of individual beetles, each carrying certain allele combinations.

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111 Types of Natural Selection become relatively more numerous and than unfavourable phenotypes. Over time, natural selection may lead to a permanent change in the genetic makeup of a population. Natural selection is always linked to phenotypic suitability in the prevailing environment so it is a dynamic process. It may favour existing phenotypes or shift the phenotypic median, as is shown in the diagrams below. The top row of diagrams below represents the population phenotypic spread before selection, and the bottom row the spread afterwards.

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Key Idea: Natural selection acts on phenotypes and results in the differential survival of some genotypes over others. It is an important cause of change in gene pools. Natural selection operates on the phenotypes of individuals, produced by their particular combinations of alleles. It results in the differential survival of some genotypes over others. As a result, organisms with phenotypes most suited to the prevailing environment are more likely to survive and breed than those with less suited phenotypes. Favourable phenotypes will Stabilising selection

Directional selection

Frequency

Frequency

Eliminated

Retained

Eliminated

Increasing birth weight

Frequency

Eliminated

Retained

Eliminated

Disruptive selection

Retained

Retained

Increasing beak size

Increasing pigmentation

Frequency

Frequency

Frequency

Two peaks

Increasing birth weight

Increasing pigmentation

Extreme variations are selected against and the middle range (most common) phenotypes are retained in greater numbers. Stabilising selection results in decreased variation for the phenotypic character involved. This type of selection operates most of the time in most populations and acts to prevent divergence of form and function, e.g. birth weight of human infants.

The adaptive phenotype is shifted in one direction and one phenotype is favoured over others. Directional selection was observed in peppered moths in England during the Industrial Revolution. In England’s current environment, the selection pressures on the moths are more typically balanced, and the proportions of each form vary regionally.

Increasing beak size

Disruptive selection favours two phenotypic extremes at the expense of intermediate forms. During a prolonged drought on Santa Cruz Island in the Galápagos, it resulted in a population of ground finches that was bimodal for beak size. Competition for the usual mediumsized seed sources was so intense that selection favoured birds able to exploit either small or large seeds.

1. Explain why fluctuating (as opposed to stable) environments favour disruptive (diversifying) selection:

2. Disruptive selection can be important in the formation of new species:

(a) Describe the evidence from the ground finches on Santa Cruz Island that provides support for this statement:

(b) The ground finches on Santa Cruz Island are one interbreeding population with a strongly bimodal distribution for the phenotypic character beak size. Suggest what conditions could lead to the two phenotypic extremes diverging further:

(c) Predict the consequences of the end of the drought and an increased abundance of medium size seeds as food:

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112 Selection for Human Birth Weight

Key Idea: Stabilising selection operates to keep human birth weight within relatively narrow constraints. Selection pressures operate on populations in such a way as to reduce mortality. For humans, selection pressures act to For this activity, you will need a sample of 100 birth weights. You can search birth records online or use the data provided in the Model Answers booklet. Group the weights into each of the 12 weight classes indicated on the graph template provided. Calculate the percentage in each weight class.

Step 3:

Graph these in the form of a histogram for the 12 weight classes (use the graphing grid provided right). Be sure to use the scale provided on the left vertical (y) axis.

Step 4:

Create a plot of percentage mortality of newborns in relation to their birth weight. Use the scale on the right y axis and data provided (below). Draw a line of best fit through the points. Weight (kg)

Mortality (%)

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3.0 3 Mortality of newborn babies 3.5to birth weight 2 related 4.0 3

Weight (kg) 4.5

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Step 2:

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The size of the baby and the diameter and shape of the birth canal are the two crucial factors in determining whether a normal delivery is possible.

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Step 1:

constrain birth weight to within narrow limits. This is a good example of stabilising selection. It is possible to document this effect by plotting birth weights for a large sample of the population. Carry out the steps below.

Mortality (%) 7

5.0 15 1.0 80 1.5 30 2.0 12 2.5 4 3.0 3 Evidence3.5 indicates that the phenotypic norm 2 is shifting. that cases 4.0Researchers estimate 3 where the 4.5baby cannot fit down 7 the birth canal have increased from 30/1000 5.0 15 in the 1960s to 36/1000 births today, indicating that there is less selection with narrow Source: Biology:against The Unitywomen & Diversity Life (4th ed), bywith Starr larger and Taggart pelvesof and babies heads.

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1. Describe the shape of the histogram for birth weights:

2. What is the optimum birth weight in terms of the lowest newborn mortality?

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3. Describe the relationship between newborn mortality and birth weight:

4. Describe the selection pressures that are operating to control the range of birth weight:

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5. How might modern medical intervention during pregnancy and childbirth have altered these selection pressures?

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113 Selection for Skin Colour in Humans

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was that skin cancer is not tied to evolutionary fitness because it affects post-reproductive individuals and cannot therefore provide a mechanism for selection. Physiological and epidemiological evidence has now shown that selection pressures on skin colour are finely balanced to produce a skin tone that regulates the effects of the sun's UV radiation on the nutrients vitamin D and folate, both of which are crucial to successful reproduction and therefore evolutionary fitness. The selection is stabilising within each latitudinal region.

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Key Idea: Skin colour is the result of a dynamic balance between two different selection pressures linked to fitness. Pigmented skin of varying tones is a feature of humans that evolved after early humans lost the majority of their body hair. However, the distribution of skin colour globally is not random; people native to equatorial regions have darker skin tones than people from higher latitudes. For many years, biologists postulated that this was because darker skins had evolved to protect against skin cancer. The problem with this explanation

Skin colour in humans: a product of natural selection

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Women also have a high requirement for calcium during pregnancy and lactation. Populations that live in the tropics receive enough ultraviolet (UV) radiation to synthesise vitamin D all year long. Those that live in northern or southern latitudes do not. In temperate zones, people lack sufficient UV light to make vitamin D for one month of the year. Those nearer the poles lack enough UV light for vitamin D synthesis most of the year (above). Their lighter skins reflect their need to maximise UV absorption (the photos show skin colour in people from different latitudes).

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Peru

Human skin colour is the result of two opposing selection pressures. Skin pigmentation has evolved to protect against destruction of folate from ultraviolet light, but the skin must also be light enough to receive the light required to synthesise vitamin D. Vitamin D synthesis is a process that begins in the skin and is inhibited by dark pigment. Folate is needed for healthy neural development in humans and a deficiency is associated with fatal neural tube defects. Vitamin D is required for the absorption of calcium from the diet and therefore normal skeletal development.

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Photo: Lisa Grey

Adapted from Jablonski & Chaplin, Sci. Am. Oct. 2002

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Long-term resident Recent immigrant Southern Africa: ~ 20-30˚S

Khoisan-Namibia

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Zulu: 1000 years ago

Northern India: ~ 10-30˚S

Aborigine

European: 300 years ago

Nuba-Sudan

Arab: 2000 years ago

West Bengal

Aptychus-Flickr

Banks of the Red Sea: ~ 15-30˚N

Rita Willaert-Flickr

Australia: ~ 10-35˚S

Tamil: ~100 years ago

The skin of people who have inhabited particular regions for millennia has adapted to allow sufficient vitamin D production while still protecting folate stores. In the photos above, some of these original inhabitants are illustrated to the left of each pair and compared with the skin tones of more recent immigrants (to the right of each pair, with the number of years since immigration). The numbered locations are on the map.

1. (a) Describe the role of folate in human physiology:

(b) Describe the role of vitamin D in human physiology:

2. (a) Early hypotheses to explain skin colour linked pigmentation level only to the degree of protection it gave from UV-induced skin cancer. Explain why this hypothesis was inadequate in accounting for how skin colour evolved:

(b) Explain how the new hypothesis for the evolution of skin colour overcomes these deficiencies:

3. Explain why, in any given geographical region, women tend to have lighter skins (by 3-4% on average) than men:

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4. The Inuit people of Alaska and northern Canada have a diet rich in vitamin D and their skin colour is darker than predicted on the basis of UV intensity at their latitude. Explain this observation:

5. (a) What health problems might be expected for people of African origin now living in northern UK?

(b) How could these people avoid these problems in their new higher latitude environment?

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114 Directional Selection in Darwin's Finches

Seal Key Idea: The effect of directional selection on a population can be verified by making measurements of phenotypic traits. Natural selection acts on the phenotypes of a population. Individuals with phenotypes that increase their fitness produce Beak depth (mm)

No. 1976 birds

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Number of birds

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The finches on the Galápagos Islands (Darwin's finches) are famous in that they are commonly used as examples of how evolution produces new species. In this activity you will analyse data from the measurement of beaks depths of the medium ground finch (Geospiza fortis) on the island of Daphne Major near the centre of the Galápagos Islands. The measurements were taken in 1976 before a major drought hit the island and in 1978 after the drought (survivors and survivors' offspring).

more offspring, increasing the proportion of the genes corresponding to that phenotype in the next generation. Many population studies have shown natural selection can cause phenotypic changes in a population relatively quickly.

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1. Use the data above to draw two separate sets of histograms:

(a) On the left hand grid draw side-by-side histograms for the number of 1976 birds per beak depth and the number of 1978 survivors per beak depth. (b) On the right hand grid draw a histogram of the beak depths of the offspring of the 1978 survivors.

2. (a) Mark the approximate mean beak depth on the graphs of the 1976 beak depths and the 1978 offspring. (b) How much has the average moved from 1976 to 1978?

(c) Is beak depth heritable? What does this mean for the process of natural selection in the finches?

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3. The 1976 drought resulted in plants dying back and not producing seed. Based on the graphs, what can you say about competition between the birds for the remaining seeds, i.e. in what order were the seeds probably used up?

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Key Idea: Disruptive selection in the finch Geospiza fortis produces a bimodal distribution for beak size. The GalĂĄpagos Islands, 970 km west of Ecuador, are home to the finch species Geospiza fortis. A study during a prolonged

Beak size vs fitness in Geospiza fortis Local fitness (2004-2006)

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Measurements of the beak length, width, and depth were combined into one single measure.

Fitness is a measure of the reproductive success of each genotype.

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Beak sizes of G. fortis were measured over a three year period (2004-2006), at the start and end of each year. At the start of the year, individuals were captured, banded, and their beaks were measured.

drought on Santa Cruz Island showed how disruptive selection can change the distribution of genotypes in a population. During the drought, large and small seeds were more abundant than the preferred intermediate seed size.

The proportion of banded individuals in the population at the end of the year gave a measure of fitness. Absent individuals were presumed dead (fitness = 0).

0.6

Higher fitness

0.4

Higher fitness

0.2

0 -2.0 -1.5 -1.0 -0.5 0 0.5 1.0 1.5 Beak size (single measure)

Fitness related to beak size showed a bimodal distribution (left) typical of disruptive selection.

2.0

A.P. Hendry et. al 2009

The presence or absence of banded individuals was recorded at the end of the year when the birds were recaptured. Beaks were measured in the recaptured birds.

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G. fortis

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115 Disruptive Selection in Darwin's Finches

Fitness showed a bimodal distribution (arrowed) being highest for smaller and larger beak sizes.

Beak size pairing in Geospiza fortis

-1.0

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S.K. Huber et al. 2007

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S.K. Huber et al. 2007

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Pairing under dry conditions Male beak size (single measure)

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Male beak size (single measure)

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S.K. Huber et al. 2007

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S.K. Huber et al. 2007

Male beak size (single measure)

A 2007 study found that breeding pairs of birds had similar beak sizes. Male and females with small beaks -1.0 -1.0 tended to breed together, and males and females with large beaks tended to breed together. Mate selection -2.0 -2.0 maintained the bimodal distribution in the population -2.0 -1.0 0 1.0 2.0 -2.0 3.0 -1.0 0 1.0 2.0 3.0 during extremely wet conditions. If beak size wasn't a Female beak size (single measure) Female beak size (single measure) factor in mate selection, the beak size would even out.

1. (a) How did the drought affect seed size on Santa Cruz Island?

(b) How did the change in seed size during the drought create a selection pressure for changes in beak size?

3. (a) Is mate selection in G. fortis random / non-random? (delete one)

(b) Give reasons for your answer:

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2. How does beak size relate to fitness (differential reproductive success) in G. fortis?

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116 Directional Selection in Moths

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Key Idea: Selection pressures on the peppered moth during the Industrial Revolution shifted the common phenotype from the grey form to the melanic (dark) form. Genetically determined melanism is a common polymorphism in animals (meaning different forms exist in the population). In the peppered moth (Biston betularia) during the Industrial Olaf Leillinger

Melanic form Genotype: MM or Mm

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The gene controlling colour in the peppered moth, is located on a single locus. The allele for the melanic (dark) form (M) is dominant over the allele for the grey (light) form (m).

Revolution, selection favoured the proliferation of dark (melanic) forms over the pale (non-melanic) forms. Intensive coal burning during this time caused trees to become dark with soot, offering melanic forms greater camouflage against predatory birds. The shift in phenotype at this time is an example of directional selection.

The peppered moth, Biston betularia, has two forms: a grey mottled form, and a dark melanic form. During the Industrial Revolution, the relative abundance of the two forms changed to favour the dark form. The change was thought to be the result of selective predation by birds. It was proposed that the grey form was more visible to birds in industrial areas where the trees were dark. As a result, birds preyed upon them more often, resulting in higher numbers of the dark form surviving.

Museum collections of the peppered moth over the last 150 years show a marked change in the frequency of the melanic form (above right). Moths collected in 1850, prior to the major onset of the Industrial Revolution in England, were mostly the grey form (above left). Fifty years later the frequency of the darker melanic forms had increased.

Grey form Genotype: mm

150

Melanic Biston betularia

90 80

100

Winter sulfur dioxide

70 60 50

50

Summer smoke

40

0

1960

1965

1970

1975

1980

1985

Summer smoke or winter sulfur dioxide (Âľg m -3)

Frequency of melanic peppered moth related to reduced air pollution

100

Frequency of melanic form of Biston betularia (%)

In the 1940s and 1950s, coal burning was still at intense levels around the industrial centres of Manchester and Liverpool. During this time, the melanic form of the moth was still very dominant. In the rural areas further south and west of these industrial centres, the occurrence of the grey form increased dramatically. With the decline of coal burning factories and the introduction of the Clean Air Act in cities, air quality improved between 1960 and 1980. Sulfur dioxide and smoke levels dropped to a fraction of their previous levels. This coincided with a sharp fall in the relative numbers of melanic moths (right).

Olaf Leillinger

Year

1. The populations of peppered moth in England have undergone changes in the frequency of an obvious phenotypic character over the last 150 years. What is the phenotypic character?

3. Describe the relationship between allele frequency and phenotype frequency:

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2. Describe how the selection pressure on the grey form has changed with change in environment over the last 150 years:

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4. The level of pollution dropped around Manchester and Liverpool between 1960 and 1985. How did the frequency of the darker melanic form change during this period?

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117 Evolution in Pocket Mice Percent reflectance / % Site

Rock type (V volcanic)

Mice coat

Rock

KNZ

V

4

10.5

ARM

V

4

9

CAR

V

4

10

MEX

V

5

10.5

TUM

V

5

27

PIN

V

5.5

11

AFT

6

30

AVR

6.5

26

WHT

8

42

8.5

15

FRA

9

39

TIN

9

39

TUL

9.5

25

POR

12

34.5

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Key Idea: The need to blend into their surroundings to avoid predation is an important selection pressure acting on the coat colour of rock pocket mice. Rock pocket mice are found in the deserts of southwestern United States and northern Mexico. They are nocturnal, foraging at night for seeds, while avoiding owls (their main predator). During the day they shelter from the desert heat in their burrows. The coat colour of the mice varies from light brown to very dark brown. Throughout the desert environment in which the mice live there are outcrops of dark volcanic rock. The presence of these outcrops and the mice that live on them present an excellent study in natural selection. The coat colour of the Arizona rock pocket mice is controlled by the Mc1r gene (a gene that in mammals is commonly associated with the production of the dark pigment melanin).

There are variations for the gene that controls coat colour. These variations are called alleles. Homozygous dominant (DD) and heterozygous mice (Dd) have dark coats, while homozygous recessive mice (dd) have light coats. The coat colour of mice in New Mexico is not related to the Mc1r gene.

107 rock pocket mice from 14 sites were collected and their coat colour and the rock colour they were found on were recorded by measuring the percentage of light reflected from their coat (low percentage reflectance equals a dark coat). The data are presented right:

BLK

V

1. (a) What are the genotypes of the dark coloured mice?

(b) What is the genotype of the light coloured mice?

CL

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2. Using the data in the table above and the grids below and on the next page, draw column graphs of the percent reflectance of the mice coats and the rocks at each of the 14 collection sites.

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DATA


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3. (a) What do you notice about the reflectance of the rock pocket mice coat colour and the reflectance of the rocks they were found on?

(b) Suggest a cause for the pattern in 3(a). How do the phenotypes of the mice affect where the mice live?

(c) What are two exceptions to the pattern you have noticed in 3(a)?

(d) How might these exceptions have occurred?

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4. What type of selection appears to be operating in each of the environments (dark and light rock)? Explain:

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5. The rock pocket mice populations in Arizona use a different genetic mechanism to control coat colour than the New Mexico populations. What does this tell you about the evolution of the genetic mechanism for coat colour?

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118 Heterozygous Advantage below, susceptibility to malaria is high in the homozygous dominant condition, but lower in the heterozygous condition. Consequently, the heterozygote has a higher fitness in malaria-prone regions. Heterozygous advantage can result in the stable coexistence of different phenotypes in a population (a state called balanced polymorphism) and can account for the persistence of detrimental alleles. The maintenance of the sickle cell mutation in malaria-prone regions is one of the few well documented examples in which the evidence for heterozygous advantage is conclusive.

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Key Idea: Heterozygous advantage is a phenomenon in which the heterozygote has a greater fitness in the prevailing environment than either of the homozygotes. Natural selection operates on phenotypes (and therefore their genotypes) in the prevailing environment. For some phenotypic conditions controlled by a single gene with two alleles, a heterozygote (an individual with two different alleles for a gene) may have a higher fitness than either of the homozygous conditions. This situation is called heterozygous advantage. In the case of the sickle cell allele outlined The sickle cell allele (HbS)

Sickle cell disease is caused by a mutation in a gene that directs the production of the human blood protein called haemoglobin. The mutant allele is known as HbS and produces a form of haemoglobin that differs from the normal form by just one amino acid in the b-chain. This small change however causes a cascade of physiological problems in people with the allele. The red blood cells containing mutated haemoglobin alter their shape to become irregular and spiky: the so-called sickle cells.

Sickle cells have a tendency to clump together and work less efficiently. In people with just one sickle cell allele plus a normal allele (the heterozygote condition HbSHb), there is a mixture of both red blood cell types and they are said to have the sickle cell trait. They are generally unaffected by the disease except in low oxygen environments (e.g. climbing at altitude). People with two HbS genes (HbSHbS) suffer severe illness and even premature death. HbS is therefore considered a lethal gene.

Areas affected by falciparum malaria

Four species of Plasmodium cause Anopheles mosquito, the malaria, but the variety caused by insect vector responsible for spreading Plasmodium. P. falciparum is the most severe

Fig. 1: Incidence of falciparum malaria

Heterozygous advantage in malarial regions

1% - 5%

5% - 10%

10% - 20%

HbHb

HbSHb

HbSHbS

All red blood cells are normal

Mixture of normal and sickle red blood cells

All red blood cells are sickle shaped

Fig. 2: Frequency of the sickle cell allele

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Falciparum malaria is widely distributed throughout central Africa, the Mediterranean, Middle East, and tropical and semi-tropical Asia (Fig. 1). It is transmitted by the Anopheles mosquito, which spreads the protozoan Plasmodium falciparum from person to person as it feeds on blood. SYMPTOMS: These appear 1-2 weeks after being bitten, and include headache, shaking, chills, and fever. Falciparum malaria is more severe than other forms of malaria, with high fever, convulsions, and coma. It can be fatal within days of the first symptoms appearing. THE PARADOX: The HbS allele offers considerable protection against malaria. Sickle cells have low potassium levels, which causes Plasmodium parasites inside these cells to die. Those with a normal phenotype are very susceptible to malaria, but heterozygotes (HbSHb) are much less so. This situation, called heterozygous advantage, has resulted in the HbS allele being present in moderately high frequencies in parts of Africa and Asia despite its harmful effects (Fig. 2). This is a special case of balanced polymorphism, called a balanced lethal system because neither of the homozygotes produces a phenotype that survives, but the heterozygote is viable.

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1. With respect to the sickle cell allele, explain how heterozygous advantage allows the retention of a generally deleterious mutation within a population:

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119 The Evolution of Antibiotic Resistance

Key Idea: Current widespread use of antibiotics has created a selective environment for the proliferation of antibiotic resistance in bacterial populations. Antibiotic resistance arises when a genetic change allows bacteria to tolerate levels of antibiotic that would normally inhibit growth. This resistance may arise spontaneously, through mutation or copying error, or by transfer of genetic material between microbes. Genomic analyses from 30,000 year

old permafrost sediments show that the genes for antibiotic resistance are not new. They have long been present in the bacterial genome, predating the modern selective pressure of antibiotic use. In the current selective environment, these genes have proliferated and antibiotic resistance has spread. For example, methicillin resistant strains of Staphylococcus aureus (MRSA) have acquired genes for resistance to all penicillins. Such strains are called superbugs.

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The evolution of drug resistance in bacteria

If the amount of antibiotic delivered is too low, or the course of antibiotics is not completed, a population of resistant bacteria develops. Within this population too, there will be variation in susceptibility. Some will survive higher antibiotic levels than others.

MRSA infections in England

Staphylococcus aureus is a common bacterium responsible for several minor skin infections in humans. MRSA is a strain that has evolved resistance to penicillin and related antibiotics. MRSA is troublesome in hospital-associated infections because patients with open wounds, invasive devices (e.g. catheters), or poor immunity are at greater risk for infection than the general public.

Mandatory reporting 2001

Hospital hygiene programmes introduced 2004

6000 4000

Voluntary reporting

2000

0

1990

Number of cases

8000

SEM of MRSA

A highly resistant population has evolved. The resistant cells can exchange genetic material with other bacteria (via horizontal gene transmission), passing on the genes for resistance. The antibiotic initially used against this bacterial strain will now be ineffective.

2010

When a person takes an antibiotic, only the most susceptible bacteria will die. The more resistant cells remain alive and continue dividing. Note that the antibiotic does not create the resistance; it provides the environment in which selection for resistance can take place.

Drug resistance genes can be transferred to non-resistant strains.

2005

Any population, including bacterial populations, includes variants with unusual traits, in this case reduced sensitivity to an antibiotic. These variants arise as a result of mutations in the bacterial chromosome.

Bacterium with greater resistance survives

2000

Mutations occur at a rate of one in every 108 replications

1995

Susceptible bacterium

Less susceptible bacterium

Year

In the UK, MRSA cases rose sharply during the early-mid 1990s, but are now declining as a result of mandatory reporting and the implementation of stringent hospital hygiene programmes.

1. What does antibiotic resistance mean?

2. (a) How does antibiotic resistance arise in a bacterial population?

(b) Describe two ways in which antibiotic resistance can become widespread:

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3. With reference to MRSA, describe the implications to humans of widespread antibiotic resistance:

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120 Insecticide Resistance and physiological mechanisms, but the underlying process is a form of natural selection, in which the most resistant organisms survive to pass on their genes to their offspring. To combat increasing resistance, higher doses of more potent pesticides are sometimes used. This drives the selection process, so that increasingly higher dose rates are required to combat rising resistance. This phenomenon is made worse by the development of multiple resistance in some pest species. Insecticides are widely used, so the development of resistance has serious environmental and economic consequences.

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Key Idea: Insect resistance to insecticide is increasing as a result of ineffectual initial applications of insecticide that only kill the most susceptible insects, leaving the more resistant ones to form a new, more resistant population. Insecticides are pesticides used to control pest insects. They have been used for hundreds of years, but their use has increased since synthetic insecticides were first developed in the 1940s. When insecticide resistance develops, the control agent will no longer control the target species. Resistance can arise through behavioural, anatomical, biochemical,

The development of resistance

Example of insecticide resistance in insects

The application of an insecticide can act as a potent selection pressure for resistance in pest insects. The insecticide acts as a selective agent, and only individuals with greater natural resistance survive the application to pass on their genes to the next generation. These genes (or combination of genes) may spread through all subsequent populations.

The Colorado potato beetle (Leptinotarsa decemlineata) is a major potato pest that was originally found living on buffalobur (Solanum rostratum ) in the Rocky mountains. It has an extraordinary ability to develop resistance to synthetic pesticides. Since the 1940s, when these pesticides were first developed, it has become resistant to more than 50 different types.

Resistant

A small proportion of the population will have the genetic makeup to survive the first application of a pesticide.

60

Cumulative number of chemicals

Susceptible

50 40 30 20

10

Change in insecticide resistance in the Colorado potato beetle

0

1950

1960

1970 1980 1990 2000 Year of reported resistance

2010

Mechanisms of resistance in insect pests

The genetic information for pesticide resistance is passed to the next generation.

The proportion of resistant individuals increases following subsequent applications of insecticide. Eventually, almost all of the population is resistant.

Insecticide resistance in insects can arise through a combination of mechanisms. ffIncreased sensitivity to an insecticide will cause the pest to avoid a treated area. ffCertain genes confer stronger physical barriers, decreasing the rate at which the chemical penetrates the insect's cuticle. ffDetoxification by enzymes within the insect's body can render the insecticide harmless. ffStructural changes to the target enzymes make the insecticide ineffective. ffNo single mechanism provides total immunity, but together they transform the effect from potentially lethal to insignificant.

1. Give two reasons why widespread insecticide resistance can develop very rapidly in insect populations: (a)

(b)

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2. Explain how repeated insecticide applications act as a selective agent for evolutionary change in insect populations:

3. With reference to synthetic insecticides, discuss the implications of insecticide resistance to human populations:

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121 Antigenic Variability in Pathogens

Key Idea: Rapid evolution in Influenzavirus is indicated by the genetically encoded changes to membrane surface proteins. These changes are adaptations to its human host. Influenza (flu) is a disease of the upper respiratory tract caused by the viral genus Influenzavirus. Three types of Influenzavirus (A, B, and C) affect humans. The most common and most virulent of these is Influenzavirus A,

(below left). Influenza viruses are constantly undergoing genetic changes by either antigenic drift or antigenic shift (below right). These genetic changes result in changes to the proteins presented on the virus's surface. The changes to the proteins prevent the human immune system from detecting the virus easily, and allow the virus to reinfect people who may already have had the flu.

Change in Influenzavirus

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Structure of Influenzavirus

Antigenic drifts are small changes in the virus which happen continually over time. Such changes mean that the influenza vaccine must be adjusted each year to include the most recently circulating influenza viruses. Antigenic shift occurs when two or more different viral strains (or different viruses) combine to form a new subtype. The changes are large and sudden and most people lack immunity to the new subtype. New influenza viruses arising from antigenic shift have caused influenza pandemics that have killed millions people over the last century. Influenzavirus A is dangerous to human health because it is capable of antigenic shift.

Viral strains are identified by the variation in their H and N surface antigens. Viruses are able to combine and readily rearrange their RNA segments, which alters the protein composition of their H and N glycoprotein spikes. The influenzavirus is surrounded by an envelope containing protein and lipids.

The genetic material is actually closely surrounded by protein capsomeres (these have been omitted here and below right in order to illustrate the changes in the RNA more clearly). The neuraminidase (N) spikes help the virus to detach from the cell after infection.

Influenzavirus subtypes in humans

Hemagglutinin (H) spikes allow the virus to recognise and attach to cells before attacking them.

The viral genome is contained on eight RNA segments, which enables the exchange of genes between different viral strains.

H1N1

Spikes

H3N2

CDC

Photo right: Electron micrograph of Influenzavirus showing the glycoprotein spikes projecting from the viral envelope

H1N2

Influenza vaccination is the primary method for preventing influenza and is 75% effective. The continual recombination of viral RNA creates different flu strains each season. For this reason, the ‘flu’ vaccination is updated annually to incorporate the antigenic properties of currently circulating strains. Three strains are chosen for each year's vaccination. Selection is based on estimates of which strains will be predominant in the following year.

1. The Influenzavirus is able to mutate readily and alter the composition of H and N spikes on its surface. (a) Why can the virus mutate so rapidly?

(b) How does this affect the ability of the immune system to recognise and respond to the virus?

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121

LINK

98

LINK

108

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2. Discuss why a virus capable of antigenic shift is more dangerous to humans than a virus undergoing antigenic drift:

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122 The Founder Effect

Key Idea: The founder effect can result in differences in allele frequencies between a parent and founder populations. If a small number of individuals from a large population becomes isolated from their original parent population, their sample of alleles is unlikely to represent the allele proportions of the parent population. This phenomenon is

called the founder effect and it can result in the colonising (founder) population evolving in a different direction to the parent population. This is particularly the case if the founder population is subjected to different selection pressures in a new environment and if the population is missing alleles that are present in the parent population.

Mainland population

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Island population Some individuals from the mainland population are carried at random to the offshore island by natural forces such as strong winds.

aa

Founders can become isolated by migration, e.g. to an island, but also by geological events, such as the formation of mountains or straits.

Aa

Aa

This population may not have the same allele frequencies as the mainland population.

Mainland population

Allele frequencies

Actual numbers

Calculate %

Colonising island population

Phenotype frequencies

Black

Allele frequencies

Actual numbers

Pale

Allele A

Allele A

Allele a

Allele a

Total

Total

Calculate %

Phenotype frequencies

Black

Pale

1. Compare the mainland population to the population which ended up on the island (use the spaces in the tables above): (a) Count the phenotype numbers for the two populations (i.e. the number of black and pale beetles). (b) Count the allele numbers for the two populations: the number of dominant alleles (A) and recessive alleles (a). Calculate these as a percentage of the total number of alleles for each population.

N AS OT SR F OO OR M US E

2. How are the allele frequencies of the two populations different?

3. Describe some possible ways in which various types of organism can be carried to an offshore island: (a) Plants:

(b) Land animals:

(c) Non-marine birds:

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174 Microgeographic isolation in garden snails

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The European garden snail (Cornu aspersum, formerly Helix aspersa) is widely distributed throughout the world, both naturally and by human introduction. However because of its relatively slow locomotion and need for moist environments it can be limited in its habitat and this can lead to regional variation. The study below illustrates an investigation carried out on two snail populations in the city of Bryan, Texas. The snail populations covered two adjacent city blocks surrounded by tarmac roads. The snails were found in several colonies in each block. Allele frequencies for the gene MDH-1 (alleles A and a) were obtained and compared. Statistical analysis of the allele frequencies of the two populations showed them to be significantly different (P << 0.05). Note: A Mann-Whitney U test was used in this instance. It is similar to a Student's t test, but does not assume a normal distribution of data (it is non-parametric).

Block B

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Block A

1

3

1

14

5

4

2

6

Road (not to sclae)

7

13

8

2

3

Source: Evolution, Vol 29, No. 3, 1975

15

9

10

4

11

5

11

12

12

10

6

7

8

13

9

Building

Colony

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Block A

MDH-1 A %

39

39

36

42

39

47

32

42

44

42

44

50

50

58

75

Block B

Snail colony (circle size is proportional to colony size).

MDH-1 A %

81

61

75

68

70

61

70

60

58

61

54

54

47

MDH-1 a %

MDH-1 a %

4. Complete the table above by filling in the frequencies of the MDH-1 a allele:

5. Suggest why these snail populations are effectively geographically isolated:

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6. Both the MDH-1 alleles produce fully operative enzymes. Suggest why the frequencies of the alleles have become significantly different.

CL

7. Identify the colony in block A that appears to be isolated from the rest of the block itself:

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123 Genetic Bottlenecks

Key Idea: Genetic bottlenecks occur when population numbers and diversity fall dramatically. Although a population's numbers may recover, its genetic diversity often does not. Populations may sometimes be reduced to low numbers by predation, disease, or periods of climatic change. These large scale reductions are called genetic (or population) bottlenecks. The sudden population decline is not necessarily selective and it may affect all phenotypes equally. Large scale catastrophic events, such as fire or volcanic eruptions, are

examples of such non-selective events. Affected populations may later recover, having squeezed through a ‘bottleneck’ of low numbers. The diagram below illustrates how population numbers may be reduced as a result of a catastrophic event. Following such an event, the gene pool of the surviving remnant population may be markedly different to that of the original gene pool. Genetic drift may cause further changes to allele frequencies. The small population may return to previous levels but with a reduced genetic diversity.

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Change in population numbers and diversity

High

High

Population numbers

Genetic diversity

Bottleneck

Low

Low

Time

Population numbers

Large population with plenty of genetic diversity.

Genetic diversity

Population crashes to a very low number and loses most of its genetic diversity.

Population grows to a large size again, but has lost much of its genetic diversity.

Tasmanian devils are the largest surviving marsupial carnivore. Although now restricted to Tasmania, devils were once found throughout mainland Australia, but became locally extinct about 3000 years ago. Genetic evidence suggests that the devils went through at least two historic population crashes, one about 30,000 years ago and another about 3000 years ago. Coupled with these historic declines are modern declines (1850 to 1950) as a result of trapping and disease. These historic population crashes are likely to be responsible for the very low diversity in the MHC I and II genes in devils. The MHC genes are important in immunity and the body's self recognition system. Low allelic diversity for MHC is implicated in the spread of devil facial tumour disease (DFTD), a contagious cancer that appeared in populations in the mid 1990s and has resulted in the loss of 80% of the devil population. The cancerous cells are transmitted when the devils fight. Ordinarily this foreign material would be recognised and destroyed by the immune system. In Tasmanian devils, the immune diversity is so low that tumours can spread without invoking an immune response. However, recent evidence also shows that some populations are developing immunity to DFTD. This may originate in individuals with MHC alleles distinctly different from the susceptible individuals.

Mike Lehmann cc 3/0

Genetic bottlenecks and low allelic diversity in Tasmanian devils

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1. Endangered species are often subjected to genetic bottlenecks. Explain how genetic bottlenecks affect the ability of a population of an endangered species to recover from its plight:

2. (a) What has been the genetic consequence of bottleneck events in the Tasmanian devil population?

(b) How has this led to increased susceptibility to disease, specifically infectious cancer?

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124 Genetic Drift

Key Idea: Genetic drift is the term for the random changes in allele frequency that occur in all populations. It has a more pronounced effect in small populations. Not all individuals, for various reasons, will be able to contribute their genes to the next generation. In a small population, the effect of a few individuals not contributing their

alleles to the next generation can have a great effect on allele frequencies. Alleles may even become lost from the gene pool altogether (frequency becomes 0%) or fixed as the only allele for the gene present (frequency becomes 100%). The random change in allele frequencies is called genetic drift.

The genetic makeup (allele frequencies) of the population changes randomly over a period of time

Generation 2

Generation 3

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Generation 1

A = 16 (53%) a = 14 (47%)

A = 15 (50%) a = 15 (50%)

Fail to locate a mate due to low poulation density

Killed

A = 12 (41%) a = 18 (59%)

Fail to locate a mate due to low poulation density

Further chance events will affect allele frequencies in subsequent generations.

This diagram shows the gene pool of a hypothetical small population over three generations. For various reasons, not all individuals contribute alleles to the next generation. With the random loss of the alleles carried by these individuals, the allele frequency changes from one generation to the next. The change in frequency is directionless as there is no selecting force. The allele combinations for each successive generation are determined by how many alleles of each type are passed on from the preceding one.

Computer simulation of genetic drift

Below are displayed the change in allele frequencies in a computer simulation showing random genetic drift. The breeding population progressively gets smaller from left to right. Each simulation was run for 140 generations.

Allele frequency ( %)

100

100

Breeding population = 2000

100

Breeding population = 200

80

80

80

60

60

60

40

40

40

20

20

20

0

0

20

40

60

80

100

120

140

0

0

20

Generations

Large breeding population Fluctuations are minimal in large breeding populations because the large numbers buffer the population against random loss of alleles. On average, losses for each allele type will be similar in frequency and little change occurs.

40

60

80

100

120

140

0

Breeding population = 20

Allele lost from the gene pool

0

20

40

60

80

100

120

140

Generations

Generations

Small breeding population Fluctuations are more severe in smaller breeding populations because random changes in a few alleles cause a greater percentage change in allele frequencies.

Very small breeding population Fluctuations in very small breeding populations are so extreme that the allele can become fixed (frequency of 100%) or lost from the gene pool altogether (frequency of 0%).

(b) Why is the effect of genetic drift more pronounced in small populations?

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1. (a) What is genetic drift?

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2. Suggest why genetic drift is an important process in the evolution of small populations:

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125 Reproductive Isolation hybrid is a cross between two species and any factor that prevents the formation of viable, fertile hybrids contributes to reproductive isolation. Single barriers may not completely stop gene flow, so most species commonly have more than one type of barrier. Single barriers to reproduction (including geographical barriers) often precede the development of a suite of reproductive isolating mechanisms (RIMs). Most operate before fertilisation (prezygotic RIMs) with postzyotic RIMs being important in preventing offspring between closely related species.

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Key Idea: Reproductive isolation maintains separate species by preventing gene flow between populations. Isolating mechanisms are barriers to successful interbreeding between species. Reproductive isolation is fundamental to the biological species concept, which defines a species by its inability to breed with other species to produce fertile offspring. Prezygotic isolating mechanisms act before fertilisation, preventing species ever mating, whereas postzygotic barriers take effect after fertilisation. Reproductive isolation prevents interbreeding (and therefore gene flow) between species. A

Geographical isolation

Geographical isolation describes the isolation of a species population (gene pool) by some kind of physical barrier, e.g. mountain range, water body, isthmus, desert, or ice sheet. Geographical barriers are not regarded as reproductive isolating mechanisms because they are not part of the species’ biology, although they are often a necessary

Malawi cichlid species

precursor to reproductive isolation in sexually reproducing populations. Geographical isolation is a frequent first step

L. Victoria

NASA Earth Observatory

in the subsequent reproductive isolation of a species. For example, geologic changes to the lake basins have been instrumental in the subsequent proliferation of cichlid fish species in the rift lakes of East Africa (right). Similarly, many Galápagos Island species (e.g. iguanas, finches) are now quite distinct from the Central and South American species from which they arose after isolation from the mainland.

L. Tanganyika

L. Malawi

Prezygotic reproductive isolating mechanisms

Lorax

Individuals of different species do not mate because they are active during different times of the day or in different seasons. Plants flower at different times of the year or even at different times of the day to avoid hybridisation (e.g. members of the orchid genus Dendrobium, which occupy the same location and flower on different days). Closely related animal species may have different breeding seasons or periods of emergence. Periodical cicadas (right) of the genus Magicicada are so named because members of each species in a particular region are developmentally synchronised, despite very long life cycles. Once their period of development underground is over (13 or 17 years depending on the species), the entire population emerges at much the same time to breed.

Bruce Marlin

Temporal isolation

Behavioural isolation

Male tree frog calling

Male frigatebird courtship display

Wing beating in male sage grouse

Damselflies mating

Complex flowers in orchids

Mechanical isolation Structural incompatibility in the reproductive organs prevents sperm transfer between individuals of different species. This is an important isolating mechanism between closely related arthropod species. Many flowering plants have coevolved with their animal pollinators and have flower structures to allow only that insect access. Structural differences in the flowers and pollen of different plant species prevents cross breeding because pollen transfer is restricted to specific pollinators and the pollen itself must be species compatible. ©2020 BIOZONE International ISBN: 978-1-98-856647-4 Photocopying Prohibited

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Albatross courtship

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Behavioural isolation operates through differences in species courtship behaviours. Courtship is a necessary prelude to mating in many species and courtship behaviours are species specific. Mates of the same species are attracted with distinctive, usually ritualised, dances, vocalisations, and body language. Because they are not easily misinterpreted, the courtship behaviours of one species will be unrecognised and ignored by individuals of another species. Birds exhibit a remarkable range of courtship displays. The use of song is widespread but ritualised movements, including nest building, are also common. Examples include the courtship bowers of bowerbirds, the elaborate displays of Galápagos frigatebirds, and the strutting and drumming displays of grouse (right).

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Postzygotic isolating mechanisms

Zebra stallion (2N = 44)

Postzygotic isolating mechanisms operate after fertilisation and are important in preventing offspring between closely related species. They involve genetic incompatibilites.

Hybrid sterility

X

Donkey jenny (2N = 62)

Karyotype of ‘Zedronkey’ offspring (2N = 53)

Hybrid sterility may occur due to the failure of meiosis to produce normal gametes in the hybrid. This can occur if the chromosomes of the two parents are different in number or structure (see the “zebronkey” karyotype, right).

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Non-viable hybrids

Mating between individuals of two species may produce a zygote, but genetic incompatibility may stop development of the zygote. Fertilised eggs often fail to divide because of mis-matched chromosome numbers from each gamete.

Y

Chromosomes contibuted by zebra stallion

X

Chromosomes contibuted by donkey jenny

Hybrid breakdown

Hybrid breakdown is a common feature of some plant hybrids. The first generation (F1) may be fertile, but the second generation (F2) are infertile or nonviable. Examples include hybrids between species of cotton (near right), species of Populus, and strains of cultivated rice Oryza. In plants, hybridisation can lead to new species formation if there is a doubling of the chromosome number during meiosis. The new plant is immediately genetically (and therefore reproductively) isolated from the parent species due to differences in chromosome number.

Cotton

Rice

1. (a) Why is a geographical barrier not considered a reproductive isolating mechanism?

(b) Identify some geographical barriers that could separate populations:

(c) Why is geographic isolation often an important first step in species formation?

2. Explain how temporal isolation stops closely related species from interbreeding:

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3. Explain why many animals have courtship displays and how this prevents breeding between species:

4. How does the structure of some orchids isolate them from other species of orchid?

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5. What is the name given to reproductive isolating mechanisms that operate before fertilisation?

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126 Allopatric Speciation

Key Idea: Allopatric speciation is the genetic divergence of a population after it becomes subdivided and isolated. Allopatric speciation refers to the genetic divergence of a species after a population becomes split and then isolated geographically. It is probably the most common mechanism

by which new species arise and has certainly been important in regions where there have been cycles of geographical fragmentation, e.g. as a result of ice expansion and retreat (and accompanying sea level changes) during glacial and interglacial periods.

Stage 1: Moving into new environments

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There are times when the range of a species expands for a variety of different reasons. A single population in a relatively homogeneous environment will move into new regions of their environment when they are subjected to intense competition (whether it is interspecific or intraspecific). The most severe form of competition is between members of the same species since they are competing for identical resources in the habitat. In the diagram on the right there is a 'parent population' of a single species with a common gene pool with regular 'gene flow' (theoretically any individual has access to all members of the opposite sex for mating purposes).

Parent population

Stage 2: Geographical isolation

Isolation of parts of the population may occur due to the formation of physical barriers, such as mountains, deserts, or stretches of water. These barriers may cut off those parts of the population that are at the extremes of the range and gene flow is prevented or rare. The rise and fall of the sea level has been particularly important in functioning as an isolating mechanism. Climatic change can leave 'islands' of habitat separated by large inhospitable zones that the species cannot traverse. Example: In mountainous regions, alpine species can populate extensive areas of habitat during cool climatic periods. During warmer periods, they may become isolated because their habitat is reduced to ‘islands’ of high ground surrounded by inhospitable lowland habitat.

Isolated population B

Parent population

River barrier prevents gene flow

Mountain barrier prevents gene flow

Isolated population A

Wetter climate

Stage 3: Different selection pressures

The isolated populations (A and B) may be subjected to quite different selection pressures. These will favour individuals with traits that suit each particular environment. For example, population A will be subjected to selection pressures that relate to drier conditions. This will favour those individuals with phenotypes (and therefore genotypes) that are better suited to dry conditions. They may for instance have a better ability to conserve water. This would result in improved health, allowing better disease resistance and greater reproductive performance (i.e. more of their offspring survive). Finally, as allele frequencies for certain genes change, the population takes on the status of a subspecies. Reproductive isolation is not yet established but the subspecies are significantly different genetically from other related populations.

Cooler climate

Subspecies B

Parent population

Drier climate

Subspecies A

Sympatric species

Stage 4: Reproductive isolation

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Species B

Parent population

River barrier disappears

Allopatric species

Mountain barrier remains

Species A

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The separated populations (isolated subspecies) undergo genetic and behavioural changes. These ensure that the gene pool of each population remains isolated and 'undiluted' by genes from other populations, even if the two populations should be able to remix (due to the removal of the geographical barrier). Gene flow does not occur. The arrows (diagram, right) indicate the zone of overlap between two species after Species B has moved back into the range inhabited by the parent population. Closely-related species whose distribution overlaps are said to be sympatric species. Those that remain geographically isolated are called allopatric species.

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1. Why do some animals, given the opportunity, move into new environments?

2. Plants are unable to move. How might plants disperse to new environments?

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3. Describe the amount of gene flow within a parent population prior to and during the expansion of a species' range:

4. Explain how cycles of climate change can cause large changes in sea level (up to 200 m):

5. (a) What kinds of physical barriers could isolate different parts of the same population?

(b) How might emigration achieve the same effect as geographical isolation?

6. (a) How might selection pressures differ for a population that becomes isolated from the parent population?

(b) Describe the general effect of the change in selection pressures on the allele frequencies of the isolated gene pool:

7. Distinguish between allopatric and sympatric species?

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8. Explain how reproductive isolation could develop in geographically separated populations (see previous pages):

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127 Speciation in Australia distribution is restricted to savannah woodland and a distinctive form is associated with each of the major woodland areas (below). The populations probably evolved from a single common ancestor, isolated by habitat changes and then unable to expand their individual distributions beyond regions of unsuitable dry habitat. Subsequently, the distribution of two of the treecreeper species has undergone a secondary range expansion (arrows), where they have extended their range beyond their region of origin into new habitat.

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Key Idea: Geographic barriers presented by inhospitable habitat are contributing to speciation in Australian treecreepers. The geographical barriers isolating populations on continents are often very different to those isolating island populations. In Australia, geographical barriers exist in the form of regions of inhospitable habitat. These create regions of preferred habitat cut off from one another. The species and subspecies of Australian treecreepers, Climacteris, are distinguishable by variations in the colour patterns of their plumage. Their

Zone of sympatry due to secondary range expansion

C. melanura melanura

C. picumnus melanota

Circles indicate a physical barrier to range expansion

C. melanura wellsi

Secondary range expansion

Photo: Aviceda

Plumage colour key Black

Brown

Rufous (reddish brown)

The Australian treecreeper Genus Climacteris

C. rufa

C. picumnus picumnus

1. (a) How many species are illustrated above? Explain your answer:

(b) Describe the distribution of these treecreeper populations in Australia:

2. Explain why there are no treecreeper populations in the central region of Australia:

3. Two species in the NE of Australia, C. melanura melanura and C. picumnus melanota, exhibit sympatric distribution. (a) What is meant by the area of sympatry in this context?

(b) What mechanisms are most likely to prevent interbreeding between these two species?

4. What is meant by secondary range expansion in the two populations above:

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5. Describe the physical barriers that have prevented the neighbouring populations from mixing (in all but one case):

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6. Predict a likely outcome to the distribution of these species, should the climate change to produce more coastal rainfall:

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128 Stages in Species Formation

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Key Idea: Speciation may occur in stages marked by increasing isolation of diverging gene pools. Physical separation is followed by increasing reproductive isolation. The diagram below shows a possible sequence of events in the origin of two new species from an ancestral population. Over time, the genetic differences between two populations

increase and the populations become increasingly isolated from each other. The isolation of the two gene pools may begin with a geographical barrier. This may be followed by progressively greater reduction in gene flow between the populations until the two gene pools are isolated and they each attain species status. A species of butterfly lives on a plateau. The plateau is covered with grassland strewn with boulders. During colder weather, some butterflies sit on the sun-heated boulders to absorb the heat, while others retreat to the lower altitude grassland to avoid the cold.

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Ancestral population

Population interbreeding

Continued mountain building raises the altitude of the plateau, separating two populations of butterflies, one in the highlands the other in the lowlands.

Population B

Population A

Gene flow common

Gene flow uncommon

In the highlands, boulder-sitting butterflies (BSBs) do better than grass-sitting butterflies (GSBs). In the lowlands, the opposite is true. BSBs only mate on boulders with other BSBs. Darker BSBs have greater fitness than light BSBs. (they can absorb more heat from the boulders). In the lowlands, light GSBs blend in with the grass and avoid predators better than darker butterflies.

Gene flow very rare

Over time, only boulder-sitting butterflies are found in the highlands and grass-sitting butterflies in the lowlands. Occasionally wind brings members of the two groups together, but if they mate, the offspring are usually not viable or have a much lowered fitness.

Race B

Time

Race A

Subspecies A

Subspecies B

Species A

Species B

Eventually gene flow between separated populations ceases as variation between the populations increases. They fail to recognise each other as members of the same species.

1. Identify the variation in behaviour in the original butterfly population:

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Separate species

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2. What were the selection pressures acting on BSBs in the highlands and GSBs in the lowlands respectively?

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129 Selective Breeding in Animals

Key Idea: Selective breeding is the process of breeding together organisms with desirable qualities (e.g. high milk yield) so the trait is reliably passed on to the next generation. Selective breeding (or artificial selection) is the process by which humans select organisms with desirable traits and breed them together so the trait appears in the next generation. The process is repeated over many generations until the characteristic becomes common. Selective breeding

often uses reproductive technologies, such as artificial insemination, so that the desirable characteristics of one male can be passed onto many offspring. This increases the rate at which the desirable trait is passed to progeny. There are problems associated with selective breeding. The gene pool becomes more constrained and some alleles may be lost. A reduction in genetic diversity decreases the ability of a species to adapt to changes in the environment.

Problems with selective breeding

All breeds of dog are members of the same species, Canis familiaris and provide an excellent example of selective breeding. The dog was the first domesticated species and, over centuries, humans have selected for desirable traits, so extensively that there are now more than 400 breeds of dogs. Until very recently, the grey wolf was considered to the ancestor of the domestic dog. However, recent (2015) genetic studies provide strong evidence that domestic dogs and grey wolves are sister groups and shared a now extinct wolf-like common ancestor, which gave rise to the dog before the agricultural revolution 12,000 years ago. Based on genetic analysis, four major clusters of ancient dog breeds are recognised. Through selective breeding, all other breeds are thought to have descended from these clusters.

Selection for a desirable phenotype can lead to a consequential emphasis of undesirable traits, often because genes for particular characteristics are linked and selection for one inadvertently selects for the other. For example, the German shepherd is a working dog, originally bred for its athleticism and ability to track targets. However in German shepherds bred to meet the specific appearance criteria of show dogs, some traits have been exaggerated so much that it causes health issues. The body shape of the show German shepherd has been selected for a flowing trot and has a pronounced sloping back. This has resulted in leg, hip, and spinal problems. In addition, selective breeding has increased the incidence of some genetic diseases such as epilepsy and blood disorders.

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The origins of domestic dogs

1: Older lineages The oldest lineages, including Chinese breeds, basenji, huskies, and malamutes.

2: Mastiff-type An older lineage that includes the mastiffs, bull terriers, boxers, and rottweilers.

3: Herding Includes German shepherd, St Bernard, borzoi, collie, corgi, pug, and greyhound

4: Hunting Most arose in Europe. Includes terriers, spaniels, poodles, and modern hounds.

Modern dog breeds exhibit a huge variety of physical and behavioural phenotypes. Selective breeding has produced breeds to meet the specific requirements of humans.

Straight-backed German shepherd

Sloped-backed German shepherd

1. (a) What is selective breeding?

(b) What are the advantages of selective breeding?

(a) Hunting large game (e.g. boar and deer):

(b) Stock control (sheep/cattle dog):

(c) Family pet (house dog):

(d) Guard dog:

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2. List the physical and behavioural traits that would be desirable (selected for) in the following uses of a dog:

3. As a group, discuss the ethical considerations of using selective breeding to "improve" dog breeds. What would it take to change breed standards to avoid health issues? Summarise your arguments and attach the summary to this page. LINK

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130 Selection in Livestock

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Key Idea: The performance of livestock can be improved by selective breeding based on measurable physical traits. Most of the economically important traits in dairy cattle (below right) are expressed only in females, but the main opportunity for selection is in males. Selection of the best bulls, combined with their worldwide use through artificial insemination and

frozen semen has seen a rapid genetic gain (i.e. the increase in performance as a result of genetic changes) in dairy cattle since the 1970s. Bulls are assigned breeding values based on the performance of their daughters and granddaughters. In this way, the bulls and cows with the best genetics can be selected to produce the next generation.

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The perfect dairy cow

Placid

Hereford,

Beef breeds: Selection is for large breeds with a high proportion of lean muscle. Desirable traits: high muscle to bone ratio, rapid growth and weight gain, hardy, easy calving, docile temperament.

High milk yield, resists mastitis

Correct conformation: avoids injury, walks and stands comfortably

Shows when on heat and conceives easily. Produces a live calf without assistance

Few metabolic disorders, maintains body condition on inexpensive rations.

Dairy breeds: Selection is based primarily on high milk production, but good health and fertility are also selected for. Desirable traits: high milk yield with good protein and fat content, fast milking speed, docile temperament, good udder characteristics (e.g. good teat placement).

Special breeds: Some cattle are bred for their suitability to climate or terrain. Scottish highland cattle (above) are a hardy, long coated breed and produce well where other breeds cannot thrive.

Breeding programs select not only for milk production, but also for fertility, udder characteristics, and good health. In addition, artificial selection can be based on milk composition, e.g. high butterfat content (a feature of the Jersey breed, above).

A2 milk, which contains the A2 form of the beta casein protein, has recently received worldwide attention for claims that its consumption lowers the risk of childhood diabetes and coronary heart disease. Selection for the A2 variant in Holstein cattle has increased the proportion of A2 milk produced in some regions. A2 milk commands a higher price than A1 milk, so there is a commercial incentive to farmers to produce it.

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2. Suggest why selective breeding has proceeded particularly rapidly in dairy cattle:

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1. Why can artificial selection effect changes in phenotype much more rapidly than natural selection?

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131 Selection and Population Change

Key Idea: Selective breeding is able to produce rapid change in the phenotypic characteristics of a population. Humans may create the selection pressure for evolutionary change by choosing and breeding together individuals with particular traits. The example of milk yield in Holstein cows (below) illustrates how humans have directly influenced the genetic makeup of Holstein cattle with respect to milk

production and fertility. Since the 1960s, the University of Minnesota has maintained a Holstein cattle herd that has not been subjected to any selection. They also maintain a herd that was subjected to selective breeding for increased milk production between 1965 and 1985. They compared the genetic merit of milk yield in these groups to that of the USA Holstein average.

Gain in genetic merit of milk yield

Fertility in holstein cows 6

1200

5

-800

-1800

-2800 -3800

Selection of sires with the desirable traits is critical to breeding programmes in dairy cattle.

-4800

-5800

4 3 2 1 0

-4

-6800

Birth year

02

99

20

96

19

93

19

90

19

87

19

84

19

81

19

78

19

75

19

72

Birth year

UMN control cows U.S. average UMN selection cows

Milk production in the University of Minnesota (UMN) herd subjected to selective breeding increased in line with the U.S. average production. In real terms, milk production per cow per milking season increased by 3740 kg since 1964. The herd with no selection remained effectively constant for milk production.

19

69

19

66

19

19

19

02

99

20

96

19

93

19

90

19

87

19

84

19

81

19

78

19

75

19

72

19

69

19

66

19

63

19

63

-2

-7800

19

Daughter pregnancy rate

200

Based on data from T.S. Sonstegard et al

Genetic merit of milk yield

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2200

UMN control cows U.S. average UMN selection cows

Along with increased milk production there has been a distinct decrease in fertility. The fertility of the University of Minnesota (UMN) herd that was not subjected to selection remained constant while the fertility of the herd selected for milk production decreased with the U.S. fertility average.

1. (a) Describe the relationship between milk yield and fertility on Holstein cows:

(b) What does this suggest about where the genes for milk production and fertility are carried?

2. What limits might this place on maximum milk yield?

3. Why is sire selection important in selective breeding, even if the characters involved are expressed only in the female?

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4. Natural selection is the mechanism by which organisms with favourable traits become proportionally more common in the population. How does selective breeding mimic natural selection? How does the example of the Holstein cattle show that reproductive success is a compromise between many competing traits?

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132 Selective Breeding in Crop Plants

Key Idea: The genetic diversity within crop varieties provides options to develop new crop plants through selective breeding. For thousands of years, farmers have used the variation in wild and cultivated plants to develop crops. Brassica oleracea is a good example of the variety that can be produced by

selectively growing plants with desirable traits. Not only are there six varieties of Brassica oleracea, but each of those has a number of sub-varieties as well. Although brassicas have been cultivated for several thousand years, cauliflower, broccoli, and brussels sprouts appeared only in the last 500 years.

Broccoli

Cauliflower

(inflorescence: cluster of flowers on a stem)

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(flower)

Cabbage

(terminal buds)

Brussels sprout (lateral buds)

Kale

(leaf)

Domestication of Brassica

At about 3750 BC in China, the cabbage was probably the first domesticated variety of its wild form to be developed. Selective breeding by humans has produced six separate vegetables from this single species: Brassica oleracea. The wild form of this species is shown in the centre of this diagram. Different parts have been developed by human selection. In spite of the enormous visible differences, if allowed to flower, all six can crosspollinate. Kale is closer to the wild type than the other related breeds.

Kohlrabi

Wild form

(stem)

(Brassica oleracea)

1. Study the diagram above and identify which part of the plant has been selected for to produce each of the vegetables:

(a) Cauliflower:

(d) Brussels sprout:

(b) Kale:

(e) Cabbage:

(c) Broccoli:

(f) Kohlrabi:

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2. Describe the feature of these vegetables that suggests they are members of the same species:

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3. What features of Brassica oleracea would humans have selected to produce broccoli?

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187

In 18th-century Ireland, potatoes were the main source of food for about 30% of the population, and farmers relied almost entirely on one very fertile and productive variety. That variety proved susceptible to the potato blight fungus which resulted in a widespread famine.

Hybrid corn varieties have been bred to minimise damage by insect pests such as corn rootworm (above). Hybrids are important because they recombine the genetic characteristics of parental lines and show increased heterozygosity and hybrid vigour.

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The number of apple varieties available now is a fraction of the many hundreds grown a century ago. Apples are native to Kazakhstan and breeders are now looking back to this centre of diversity to develop apples resistant to the bacterial disease that causes fire blight.

4. (a) Describe a phenotypic characteristic that might be desirable in an apple tree:

(b) Outline how selective breeding could be used to establish this trait in the next generation:

5. (a) Explain why genetic diversity might decline during selective breeding for particular characteristics:

(b) With reference to an example, discuss why retaining genetic diversity in crop plants is important for food security:

6. Cultivated American cotton plants have a total of 52 chromosomes (2N = 52). In each cell there are 26 large chromosomes and 26 small chromosomes. Old World cotton plants have 26 chromosomes (2N = 26), all large. Wild American cotton plants have 26 chromosomes, all small. How might cultivated American cotton have originated from Old World cotton and wild American cotton:

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7. The Cavendish is the variety of banana most commonly sold in world supermarkets. It is seedless, sterile, and under threat of extinction by Panama disease Race 4. Explain why Cavendish banana crops are so endangered by this fungus:

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8. Why is it important to maintain the biodiversity of wild plants and ancient farm breeds?


133 Breeding Modern Wheat

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in wheat cultivars is produced by crossing inbred lines and selecting for desirable traits in the progeny, which can now be identified using genetic techniques such as marker assisted selection. This is an indirect selection process where a trait of interest is selected on the basis of a marker linked to it. Increasingly, research is focused on enhancing the genetic diversity of wheat to provide for future crop development. With this in mind, there is renewed interest in some of the lower yielding, ancient wheat varieties, which possess alleles no longer present in modern inbred varieties.

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Key Idea: Modern wheat evolved as a result of two natural hybridisation events and the doubling of its chromosomes. Wheat has been cultivated for more than 9000 years and has undergone many genetic changes during its domestication. The evolution of modern bread wheat from its wild ancestors (below) involved two natural hybridisation events, accompanied by polyploidy. Once wheat became domesticated, selective breeding emphasised characteristics such as high protein (gluten) content, high yield, and pest and disease resistance. Hybrid vigour (improved characteristics)

The evolution and domestication of wheat

X

Wild einkorn AA

Einkorn AA

Wild einkorn becomes domesticated in the Middle East. There are slight changes to phenotype but not chromosome number.

Ancient cereal grasses had heads which shattered easily so that the seeds were widely scattered. In this more primitive morphology, the wheat ear breaks into spikelets when threshed, and milling or pounding is needed to remove the hulls and obtain the grain. Cultivation and repeated harvesting and sowing of the grains of wild grasses led to domestic strains with larger seeds and sturdier heads. Modern selection methods incorporate genetic techniques to identify and isolate beneficial genes, e.g. the RHt dwarfing gene, which gave rise to shorter stemmed modern wheat varieties.

X

Wild grass BB

Emmer wheat Goat grass AABB DD

Common wheat AABBDD

A sterile hybrid between einkhorn and wild grass undergoes a chromosome doubling to create fertile emmer wheat.

A sterile hybrid between emmer wheat and goat grass undergoes a chromosome doubling to create fertile common wheat.

Modern bread wheat has been selected for its non-shattering heads, high yield, and high gluten (protein) content. The grains are larger and the seeds (spikelets) remain attached to the ear by a toughened rachis during harvesting. On threshing, the chaff breaks up, releasing the grains. Selection for these traits by farmers might not necessarily have been deliberate, but occurred because these traits made it easier to gather the seeds. Such 'incidental' selection was an important part of crop domestication. Hybrid vigour in cultivars is generated by crossing inbred lines.

Durum wheat is a modern variety developed by selective breeding of the domesticated emmer wheat strains. Durum (also called hard wheat) has large firm kernels with a high protein content. These properties make it suitable for pasta production. As with all new wheat varieties, new cultivars are produced by crossing two lines using hand pollination, then selfing or inbreeding the progeny that combine the desirable traits of both parents. Progeny are evaluated for several years for the traits of interest, until they can be released as established varieties or cultivars.

(a) (b) (c)

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1. Describe three phenotypic characteristics that would be desirable in a wheat plant:

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2. How have both natural events and selective breeding contributed to the high yielding modern wheat varieties?

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134 KEY TERMS AND IDEAS: Did You Get It?

1. Test your vocabulary by matching each term to its definition, as identified by its preceding letter code.

adaptation

allopatric fitness

A The condition of having a chromosome complement of more than 2N (e.g. 3N) B The observable characteristics in an organism. C A heritable characteristic of a species that equips it for survival and reproductive success in its environment. D Speciation in which the populations are physically separated. A term for physically separated populations.

genetic drift

E The differences between individuals in a population as a result of genes and environment.

mutation

F The process by which favorable heritable traits become more common in successive generations.

natural selection

G The collective group of genes in a population.

phenotype

H Random changes in allele frequency between generations as a result of the different contributions of individuals to the alleles in the gene pool of the next generation.

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polyploidy

I A measure of an individual's relative genetic contribution to the next generation as a result of its combination of traits.

variation

J A change in the base sequence of DNA; the ultimate source of new alleles.

2. Outline the features of the four factors involved in evolutionary change in a population: (i) Species population:

(ii) Genetic variation:

(iii) Competition:

(iv) Proliferation:

3. For the DNA sequence G G T C T C C G T A A T A T T show the effect of the following by writing the new DNA sequence (note: position one in the sequence starts with the bold G on the left hand side):

(a) A deletion of the T at position 5:

(b) A substitution of C at position 7 with A:

(c) An inversion of the bases from positions 5 - 10:

(d) A deletion of the bases from positions 10 - 12:

(a) Directional selection:

(b) Stabilising selection:

(c) Disruptive selection:

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4. Explain the characteristics of each of the following types of natural selection and state when each might operate:

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Changes in biodiversity over time

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Unit 4 Outcome 1

Key terms

The history of life on Earth

adaptation

Key knowledge

adaptive radiation

analogous structure

background extinction rate

Activity number

c

1

Understand the biochemical evidence (the universality of DNA, amino acids, and protein structures) for the common ancestry of living organisms. Recall the evidence for the endosymbiotic origin of mitochondria and chloroplasts.

c

2

Outline the significant events in the history of life on Earth including the rise of multicellular organisms, the establishment of terrestrial life, and the rise of the mammals. What were the possible triggers for these major shifts in biodiversity?

biogeographical evidence

136

136 137

chronometric dating (= absolute dating) common ancestor

comparative anatomy continental drift

convergent evolution divergent evolution (=cladogenesis) evo-devo

Evidence of evolution

evolution

Activity number

Key knowledge

extinction

RA

fossil

c

3

Understand that the evidence for evolution comes from many disciplines.

fossil record

c

4

138 - 140

geologic time scale

Using examples, describe the palaeontological evidence for evolution. Describe types of fossils and explain how fossils are formed and dated.

c

5

Explain the significance of transitional fossils. Using examples, describe the trends that fossils indicate in the evolution of related groups.

141 - 143

homologous structure

c

6

Explain how the geographical distribution of living and extinct organisms (biogeography) provides evidence of dispersal from a point of origin.

144 - 147

c

7

Recall how founder populations may establish and evolve in isolation from their parent population. Describe features of typical oceanic island colonisers and give examples of where this has been important in the evolution of new species.

146 147

c

8

Explain how comparisons of structural morphology and physiology have contributed to our understanding of evolutionary relationships.

141 - 144 148

c

9

Explain how evolutionary developmental biology (evo-devo) has provided strong evidence for the mechanisms of evolution and the evolution of novel forms.

c

10

mass extinction

phylogeny

relative dating rock strata

transitional fossil

Appreciate that evidence for evolution is not confined to the past. There are many documented examples of evolution occurring in the present day.

Patterns of evolution

149 165

117 - 121

Activity number

Key knowledge

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radiometric dating

c

11

Recognise patterns of biological change over geological time to include divergent evolution, convergent evolution, and mass extinction. Appreciate how evolution has resulted in a great diversity among living organisms.

150 151

c

12

Explain adaptive radiation as a type of divergent evolution in which there is a rapid diversification of an ancestral group into many different species. Describe homologous structures arising as a result of divergent evolution.

150 - 153

c

13

Explain convergent evolution as the evolution of similar adaptations in unrelated taxa. Convergent evolution is a response to similar selection pressures producing similar solutions from different points of origin (analogous structures).

154

c

14

Describe the role of extinction in Earth's biological history. Distinguish clearly between background extinction rates and mass extinction. Identify the major mass extinctions and discuss their possible causes.

155

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palaeontological evidence

135


191

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135 Evidence for Evolution changes must be passed on to the next generation (i.e. be inherited). The evidence for evolution comes from many diverse branches of science and includes evidence from both past and present populations. Drawing on evidence from a number of scientific disciplines helps to build a robust explanation for the evolutionary history of taxa.

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Key Idea: Evidence for the fact that populations evolve comes from many fields of science. Evolution is simply the heritable genetic changes occurring in a population over time. There are two important points to take from this definition. The first is that evolution refers to populations, not individuals. The second is that the

Comparative anatomy

Geology

Comparative anatomy examines the similarities and differences in the anatomy of different species. Similarities in anatomy (e.g. the bones forming the arms in humans and the wings in birds and bats) indicate descent from a common ancestor.

Geological strata (the layers of rock, soil, and other deposits such as volcanic ash) can be used to determine the relative order of past events and therefore the relative dates of fossils. Fossils in lower strata are older than fossils in higher (newer) strata, unless strata have been disturbed.

DNA comparisons

Cytochrome c

DNA can be used to determine how closely organisms are related to each other. The greater the similarities between the DNA sequences of species, the more closely related the species are.

Protein evidence

Similarities (and differences) between proteins provides evidence for determining shared ancestry. Fewer differences in amino acid sequences reflects closer genetic relatedness.

EVOLUTION

Fossil record

Developmental evidence

Fossils, like this shark's tooth (left) are the remains of longdead organisms. They provide a record of the appearance and extinction of organisms.

The study of developmental processes and the genes that control them gives insight into evolutionary processes. This field of study is called evolutionary developmental biology (evo-devo).

The geographical distribution of living and extinct organisms provides evidence of common ancestry and can be explained by speciation, extinction, and continental drift. The biogeography of islands, e.g the GalĂĄpagos Islands, provides evidence of how species evolve when separated from their ancestral population on the mainland.

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Radiometric dating techniques (such as carbon dating) allow scientists to determine an absolute date for a fossil by dating it or the rocks around it. Absolute dating has been used to assign ages to strata, and construct the geological time scale.

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Biogeography

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Chronometric dating

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135

REFER


192

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136 Descent and Common Ancestry how all life on Earth is related. These newer methods have enabled scientists to clarify the origin of the eukaryotes and to recognize two prokaryote domains. The universality of the genetic code and the similarities in the molecular machinery of all cells provide powerful evidence for a common ancestor to all life on Earth.

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Key Idea: Molecular studies have enabled scientists to clarify the earliest beginnings of the eukaryotes. Such studies provide powerful evidence of the common ancestry of life. Traditionally, the phylogeny (evolutionary history) of organisms was established using morphological comparisons. In recent decades, molecular techniques involving the analysis of DNA, RNA, and proteins have provided more information about

There is a universal genetic code

DNA encodes the genetic instructions of all life. The form of these genetic instructions, called the genetic code, is effectively universal, i.e. the same combination of three DNA bases code for the same amino acid in almost all organisms. The very few exceptions in which there are coding alternatives are restricted to some bacteria and to mitochondrial DNA.

Domain Bacteria

Cyanobacteria

Proteobacteria (many pathogens)

Other bacteria

Rocky Mountain Laboratories, NIAID, NIH

Hyperthermophillic bacteria

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LINK

45

Chloroplasts have a bacterial origin Cyanobacteria are considered to be the ancestors of chloroplasts. The evidence for this comes from similarities in the ribosomes and membrane organisation , as well as from genomic studies. Chloroplasts were acquired independently of mitochondria, from a different bacterial lineage, but by a similar process. LINK

56

Mitochondria have a bacterial origin

Evidence from mitochondrial gene sequences, ribosomes, and protein synthesis indicate that mitochondria have a prokaryotic origin. Mitochondria were probably symbiotic inclusions in an early eukaryotic ancestor.

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Bacteria lack a distinct nucleus and cell organelles. Features of the cell wall are unique to bacteria and are not found among archaea or eukaryotes. Typically found in less extreme environments than archaea.

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EII

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1. Identify three features of the metabolic machinery of cells that support a common ancestry of life: (a) (b) (c)

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2. Suggest why scientists believe that mitochondria were acquired before chloroplasts:

Eukarya (the eukaryotes) are characterised by complex cells with organelles and a membrane-bound nucleus. This domain contains four of the kingdoms recognised under a traditional scheme.

Archaea resemble bacteria but membrane and cell wall composition and aspects of metabolism are very different. They live in extreme environments similar to those on primeval Earth.

Domain Eukarya

Animals

Fungi

Plants

Algae

Domain Archaea

Ciliates

RCN

loroplasts Bacteria that gave rise to ch

Bacteria that g

ave rise to mitochondria

Eukaryotes have linear chromosomes Eukaryotic cells all have large linear chromosomes (above) within the cell nucleus. The evolution of linear chromosomes was related to the appearance of mitosis and meiosis.

Xiangyux (PD)

Eukaryotes have an archaean origin

Living systems share the same molecular machinery

EII

In all living systems, the genetic machinery consists of selfreplicating DNA molecules. Some DNA is transcribed into RNA, some of which is translated into proteins. The machinery for translation (left) involves proteins and RNA. Ribosomal RNA analysis support a universal common ancestor.

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Last Universal Common Ancestor (LUCA)

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Archaea superficially resemble bacteria but similarities in their molecular machinery (RNA polmerase and ribosome proteins) show that they are more closely related to eukaryotes.


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137 Changes in the Earth's Biodiversity Over Time

Key Idea: Life on Earth originated about 4100 million years ago, but complex life evolved only about 600 million years. Life forms on Earth originally arose from primitive cells living some 4100 million years ago in conditions quite different to those on Earth today. The earliest fossil records of living things show only simple cell types. It is thought that the first cells arose as a result of evolution at the chemical level in a ‘primordial soup’ (a rich broth of chemicals in a warm pool of

water, perhaps near a volcanic vent). Life appears very early in Earth’s history, but did not evolve beyond the single cell stage until much later (about 600 mya). This would suggest that the evolution of complex life forms required greater hurdles to be overcome. The build up of free atmospheric oxygen, released as a by-product of oxygenic photosynthesis, was important for the evolution of eukaryotes and paved the way for the evolution of multicellular life.

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Time line of Earth's biodiversity

First evidence of First photosynthetic First eukaryotes life (Greenland) organisms First bacterial Build up of oxygen First multicellular fossils (Australia) in atmosphere organisms Origin of life

Formation of the Earth MIllions of years ago 4500

4000

3500

Hadean

3000

2500

2000

Archean

First land plants

First land animals

400

Ordovician

300

Silurian Devonian

Phanerozoic

200

Carboniferous

Permian

Triassic

Jurassic

Cenozoic

Cretaceous

Evolution of grasses

40

30

Eocene

20

10

Neogene

Oligocene

Period

Modern humans

Palaeogene

Palaeocene

Era

Quaternary

50

Aeon

100

Mesozoic

Diversification of mammals

60

0

500

First First birds mammals First flowering Extinction of plants First reptiles Dinosaurs dinosaurs

Palaeozoic

Cambrian

1000

Proterozoic

First amphibians

500

1500

Miocene

Pliocene

Period Epoch

Pleistocene Holocene

1. (a) What was the significance of the buildup of free oxygen in the atmosphere for the evolution of life?

(b) How long did it take for free oxygen to build up in the atmosphere?

2. Explain how multicellular life evolved:

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3. What were the earliest land animals and what circumstances may have caused them to come on to the land?

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Stromatolites (such as the ones shown left from Shark Bay, Western Australia), represent some of the most ancient living things on Earth. Few examples exist today, but fossil remains can be dated back to 3.7 billion years ago. Stromatolites are rock like structures formed from the accretion of sediment by microorganisms, especially cyanobacteria. Cyanobacteria are photosynthetic bacteria. Ancient representatives of cyanobacteria are thought to be responsible for the production of oxygen in the atmosphere after evolving oxygenic photosynthesis, a event called the Great Oxygenation Event. The production of oxygen caused the extinction of many anaerobic bacteria but eventually led to the rise of multicellular life forms.

RA

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195

Fossil jellyfish

Ashley Dace

46 cm

Multicellular organisms arose soon after the evolution of eukaryotes. Multicellularity was a major evolutionary event as it allowed organisms to diversify the tissues and cells of their bodies to perform specialised tasks. The origin of multicellularity is much debated but one hypothesis is that unicellular organisms began to associate together (e.g. cyanobacteria stick together after binary fission and form long chains called filaments). Different cells in the group produced molecules useful to others and so the group benefited by staying together. As the different cell lines became more dependent on others for certain molecules, a greater need to remain together also developed (in low nitrogen conditions, some of the cells in filamentous cyanobacteria transform into nitrogen-fixing cells, and this benefits the other cells in the filament). Animals may have ventured onto land before plants, with evidence suggesting they did so about 530 million years ago. Plants may have arrived soon after, as early as 500 million years ago. The earliest land animals were invertebrates, perhaps similar to horseshoe crabs, which come ashore to lay eggs on the sand. Strategies like this would have been an advantage at a time when there were no land animals to eat the eggs. Similarly, some of the first excursions onto land may have been to take refuge from aquatic predators. The earliest terrestrial plants had no vascular tissue, like mosses and liverworts today. Vascular plants (e.g. ferns) did not appear until about 425 million years ago.

Arthropleura (extinct millipede) tracks

Flowering plants (angiosperms) are the most successful terrestrial plants. With at least 350,000 species, they make up 90% of all living plant species. Flowering plants first appeared about 160 million years ago. They began to diversify rapidly about 120 million years ago. The evolution of the flower helped to make sexual reproduction far more efficient. Flowers attracted insects (and later birds) with the use of colours and rewards. The insects and birds then spread pollen from flower to flower. This system has become so successful that many insects and birds now rely on flowers for food and plants rely on their pollinators for reproduction. Genetic evidence suggests the evolution of flowers was linked to at least two rounds of whole genome duplication, which might explain why angiosperms appeared suddenly in the fossil record.

Udo SchrĂśter CC 2.0

Mammalian evolution can be traced back to the Carboniferous period with the appearance of the synapsids (e.g. Dimetrodon), one of the two major clades of tetrapod vertebrates (the other clade gave rise to the reptiles and birds). It was not until the Triassic period that the first true mammals appeared. The monotremes (egg laying mammals) appeared about 210 million years ago. They are represented today by the platypus and echidna. Marsupials and placentals probably split about 125 million years ago. Today marsupials are found almost exclusively in Central and South America and Australia. While there are 334 species of marsupials, there are nearly 4000 species of placental mammal. It wasn't until the extinction of the dinosaurs at the end of the Cretaceous that mammals were able to fully diversify into the wide range we see today.

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4. Explain why the evolution of flowers was an advantage to plants:

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5. Explain why mammals did not diversify until the Palaeocene epoch, even though they first appeared in the Triassic period.


138 Fossil Formation

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196

requires the normal processes of decay to be permanently arrested. This can occur if the organism's remains are isolated from the air or water and decomposing microbes are prevented from breaking them down. Fossils provide a record of the appearance and extinction of organisms, from species to whole taxonomic groups. Once this record is calibrated against a time scale (by using a broad range of dating techniques), it is possible to build up a picture of the evolutionary changes that have taken place.

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Key Idea: Fossils are the remains of long-dead organisms that have escaped decay and have, after many years, become part of the Earth’s crust. A fossil may be the preserved remains of the organism itself, the impression of it in the sediment (a mould), or marks made by it during its lifetime (trace fossils). For fossilisation to occur, rapid burial of the organism is required (usually in waterborne sediment). This is followed by chemical alteration, whereby minerals are added or removed. Fossilisation Fossilisation best occurs when an organism dies in a place where sediment can be laid down relatively quickly. This is often an aquatic environment, e.g. an estuary, but it can be caused by rapid burial, e.g. by a landslide or volcanic ash.

After death, the flesh may rot or be scavenged, but hard materials, usually bones and teeth, are able to remain long enough for burial.

Soft material such as the cartilaginous skeletons of sharks don't fossilise well. Often the only remains are their teeth.

After burial, the bones are subjected to pressure. Minerals in the surrounding sediments move into the bones and replace the minerals in them.

Erosion of the sediments exposes the fossils on the surface.

1. Briefly describe how a fossil forms:

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2. Explain why the rapid burial of an organism is important in the formation of fossils:

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138 139 140 141

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3. Explain why the fossil record is biased towards marine organisms with hard parts:

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197 Modes of preservation ff Silicification: Silica from weathered volcanic ash is gradually incorporated into partly decayed wood (also called petrification). ff Phosphatisation : Bones and teeth are preserved in phosphate deposits. ff Pyritisation : Iron pyrite replaces hard remains of the dead organism. ff Tar pit: Animals fall into and are trapped in mixture of tar and sand. ff Trapped in amber: Gum from conifers traps insects and then hardens. ff Limestone: Calcium carbonate from the remains of marine plankton is deposited as a sediment that traps the remains of

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other sea creatures.

Bark

All photos: RA

Ray structure

Ants

Polished amber

Growth rings largely destroyed

Insects in amber: The fossilised resin or gum produced by some ancient conifers trapped these insects (including the ants visible in the enlargement) about 25 million years ago (Madagascar).

Petrified wood: A cross-section of a limb from a coniferous tree (Madagascar).

Sand and tar matrix

Shell

Stone interior

Wing bones

Bird bones: Fossilised bones of a bird that lived about 5 million years ago and became stuck in the tar pits at la Brea, Los Angeles, USA.

Ammonite: This ammonite still has a layer of the original shell covering the stone interior, Jurassic (Madagascar).

Mould: This impression of a scallop-like is all that is left after the original shell material was dissolved after fossilisation.

Shell and chambers replaced by iron pyrite

Soft mudstone

Carbon

Impressions of leaf veins

Compression fossil: This fern frond shows traces of carbon and wax from the original plant, Carboniferous (USA).

Sub-fossil: Leaf impression in soft mudstone (can be broken easily with fingers) with some of the remains of the leaf still intact (a few thousand years old, New Zealand).

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Cast: This ammonite has been preserved by a process called pyritisation, late Cretaceous (Charmouth, England).

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4. Explain why finding organisms trapped in amber can be a particularly useful find:


139 Relative Dating and the Fossil Record

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198

Key Idea: Fossils found in rock at the bottom of a rock profile are older than those found at the top of the rock file. Relative dating establishes the sequential (relative) order of past events in a rock profile, but it cannot provide an absolute date for an event. Each rock layer (stratum) is unique in terms of the type of rock (sedimentary or volcanic) and the

type of fossils it contains. Rock layers (strata) are arranged in the order that they were deposited, with the oldest layers at the bottom (unless disturbed by geological events). This is called the law of superposition. Strata from widespread locations with the same fossils or characteristics can thus be correlated, even when their absolute date is unknown.

Profile with sedimentary rocks containing fossils Recent fossils are found in more recent sediments

African and Indian elephants have descended from a diverse group known as proboscideans (named for their long trunks). The first pig-sized, trunkless members of this group lived in Africa 40 million years ago. From Africa, their descendants invaded all continents except Antarctica and Australia. As the group evolved, they became larger, an effective evolutionary response to deter predators. Examples of extinct members of this group are illustrated below:

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Ground surface

The fossil record of proboscidea

Youngest sediments

The more recent the layer of rock, the more resemblance there is between the fossils found in it and living forms. Many extinct species

The number of extinct species is far greater than the number of species living today.

Fossil types differ in each stratum

Woolly mammoth Pleistocene, north of the Arctic Circle

Fossils in any given layer of sedimentary rock are generally quite different to fossils in other layers.

Stegodon Pliocene-Pleistocene, Asia, Africa

Deinotherium Miocene-Pleistocene, Asia, Africa

More primitive fossils are found in older sediments

Gomphotherium

Phyla are represented by more generalised forms in the older layers, and not by specialised forms (such as those alive today).

Oldest sediments

Miocene, Europe, Africa

What is relative dating?

In the rocks marking the end of one geologic period, it is common to find many new fossils that become dominant in the next. Each geologic period had an environment very different from those before and after. Their boundaries coincided with drastic environmental changes and the appearance of new niches. New selection pressures resulted in new adaptive features as species responded to the changes. An absolute age can be assigned to fossils, usually by dating the rocks around them. Most often, this involves radiometric dating (e.g. radiocarbon, K-Ar).

1. Explain the importance of fossils in relative dating:

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139 140

Patybelodon Middle Miocene, Northern Asia, Europe, Africa

• Modern day species can be traced: The evolution of many present-day species can be very well reconstructed. For instance, the evolutionary history of the modern elephants is exceedingly well documented for the last 40 million years. The modern horse also has a well understood fossil record spanning the last 50 million years. • Fossil species are similar to but differ from today's species: Most fossil animals and plants belong to the same major taxonomic groups as organisms living today. However, they do differ from the living species in many features.

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New fossil types mark changes in environment

Ambelodon Middle Miocene, North America

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Relative dating is a way to determine the relative order of past events without necessarily determining absolute (chronometric) age. The same rocks and fossils can then be used to correlate stratigraphic records in different places. Material that can't be dated using absolute methods can therefore be correlated with the same material elsewhere for which an absolute date may be available.

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199 Rock profile at location 1

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Trilobite fossil Dated at 375 million years

A

Fossils are embedded in the different layers of sedimentary rock

Rock profile at location 2

B

I

C

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J K

D

L

E

M

A distance of 67 km separates these rock formations

F

N

O

G

H

The questions below relate to the diagram above, showing a hypothetical rock profile from two locations separated by a distance of 67 km. There are some differences between the rock layers at the two locations. Apart from layers D and L which are volcanic ash deposits, all other layers comprise sedimentary rock.

2. Assuming there has been no geologic activity (e.g. tilting or folding), state in which rock layer (A-O) you would find:

(a) The youngest rocks at location 1:

(c) The youngest rocks at location 2:

(b) The oldest rocks at location 1:

(d) The oldest rocks at location 2:

3. (a) State which layer at location 1 is of the same age as layer M at location 2:

(b) Explain the reason for your answer above:

4. The rocks in layer H and O are sedimentary rocks. Explain why there are no visible fossils in these layers:

5. (a) State which layers present at location 1 are missing at location 2:

(b) State which layers present at location 2 are missing at location 1: 6. Using radiometric dating, the trilobite fossil was determined to be approximately 375 million years old. The volcanic rock layer (D) was dated at 270 million years old, while rock layer B was dated at 80 million years old. Give the approximate age range (i.e. greater than, less than, or between given dates) of the rock layers listed below:

(a) Layer A:

(d) Layer G:

(b) Layer C:

(e) Layer L:

(c) Layer E:

(f) Layer O:

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7. Suggest why gaps in the fossil record can make it difficult to determine an evolutionary history?


200

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140 Absolute Dating

Key Idea: Absolute dating uses methods including radioisotope analysis that give an absolute time since death. Radiometric dating methods allows an absolute date to be assigned to fossils, usually by dating the rocks around the fossils. Multiple dating methods for samples provides cross-referencing, which gives a high degree of confidence

Usable dating range (years)

Datable materials

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Dating method

in a given date. Absolute, or chronometric, dating methods most often involve radiometric dating (e.g. radiocarbon, potassium-argon, fission track), which relies on the radioactive decay of an element. Non-radiometric methods (e.g. treerings, paleomagnetism) can be used in certain specific circumstances.

Methods using radioisotopes

100 million

10 million

Fission track: Uranium (235U) sometimes undergoes spontaneous fission, and the subatomic particles emitted leave tracks through the mineral.

1 million

100,000

1000 (Log scale)

10,000

Pottery, glass, and volcanic minerals.

Pottery amphora

Radiocarbon (14C): Measures the loss of the isotope carbon-14, taken up by an organism when it was alive, within its fossilised remains.

Wood, shells, peat, charcoal, bone, animal tissue, calcite, soil.

Bone

Potassium/Argon (K/Ar): Measures the decay of potassium-40 to argon-40 in volcanic rocks that lie above or below fossil bearing strata.

Uranium series: Measures the decay of the two main isotopes of uranium (235U and 238U) into thorium (230Th) and another isotope (234U) respectively.

Volcanic rocks and minerals.

Basalt

Marine carbonate, coral, mollusc shells

Mollusc shell

Non-isotopic methods

Palaeomagnetism: Shows the alignment of the Earth’s magnetic field when the rock sample was last heated above a critical level. Thermoluminescence: Measures the light emitted by a sample of quartz and/or zircon grains that has been exposed to sunlight or fire in the distant past.

Rocks that contain iron-bearing minerals

Hematite

Ceramics, quartz, feldspar, carbonates

Flint (quartz)

Electron spin resonance (ESR): Measures the microwave energy absorbed by samples previously heated or exposed to sunlight in the distant past.

Amino acid racemisation: Measures the gradual conversion of left- to right-handed amino acid isomers in the proteins preserved in organic remains.

Shark tooth

Burnt flints, cave sediments, bone, teeth, loess (windblown deposits)

Organic remains

Pollen

Varve: Measures the annually deposited sediment layers (varves) found in many lakes.

Mainly glacial lakes

Tree-ring: Measures annual growth rings of trees (can be cross referenced with 14C dates).

Trees, timber from buildings, ships

Tree rings

1. Examine the diagram above and determine the approximate dating range (note the logarithmic time scale) and datable materials for each of the methods listed below: Dating range

(a) Potassium-argon method:

(b) Radiocarbon method:

(c) Tree-ring method:

(d) Thermoluminescence:

Datable materials

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2. When the date of a sample has been determined, it is common practice to express it in the following manner: Example: 1.88 ± 0.02 million years old. Explain what the ± 0.02 means in this case:

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141 Transitional Fossils fossils include horses, whales, and Archaeopteryx and other non-avian feathered dinosaurs (below). Archaeopteryx was a transitional form between non-avian dinosaurs and birds. Archaeopteryx was crow-sized (50 cm length) and lived about 150 million years ago. It is regarded as the first primitive bird and had a number of birdlike (avian) features, including feathers. However, it also had many non-avian features, which it shared with theropod dinosaurs of the time.

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Key Idea: Transitional fossils show intermediate states between two different, but related, groups. They provide important links in the fossil record Transitional fossils are fossils with a mixture of features found in two different, but related, groups. Transitional fossils provide important links in the fossil record and provide evidence to support how one group may have given rise to the other by evolutionary processes. Important examples of transitional

Non-avian features

Avian features

Forelimb has three functional fingers with grasping claws

Vertebrae are almost flat-faced

Lacks the reductions and fusions present in other birds

Impressions of feathers attached to the forelimb

Breastbone is small and lacks a keel True teeth set in sockets in the jaws

Belly ribs

The hind-limb girdle is typical of dinosaurs, although modified

Incomplete fusion of the lower leg bones

Long, bony tail, shared with other dinosaurs of the time

KPryor

Impressions of feathers attached to the tail

Suggested reconstruction of Archaeopteryx based on fossil evidence.

1. (a) What is a transitional fossil?

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(b) Why are transitional fossils important in understanding evolution?

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142 The Evolution of Whales

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Key Idea: The evolution of whales is well documented in the fossil record, with many transitional forms recording the shift from a terrestrial to an aquatic life. The evolution of modern whales from an ancestral land mammal is well documented in the fossil record. The fossil

record of whales includes many transitional forms, which has enabled scientists to develop an excellent model of whale evolution. The evolution of the whales (below) shows a gradual accumulation of adaptive features that have equipped them for life in the open ocean.

Baleen whales: Toothless whales, which have a comb-like structure (baleen) in the jaw. Baleen is composed of the protein keratin and is used to filter food from the water. Examples: blue whale, humpback whale.

Orca

Robert Pittman - NOAA

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Modern whales are categorised into two broad suborders based on the presence or absence of teeth. Toothed whales: These have full sets of teeth throughout their lives. Examples: sperm whale and orca.

Humpback whale

50 mya Pakicetus

Pakicetus was a transitional species between carnivorous land mammals and the earliest true whales. It was mainly land dwelling, but foraged for food in water. It had four, long limbs. Its eyes were near the top of the head and its nostrils were at the end of the snout. It had external ears, but they showed features of both terrestrial mammals and fully aquatic mammals.

45 mya Rhodocetus

Rhodocetus was mainly aquatic (water living). It had adaptations for swimming, including shorter legs and a shorter tail. Its eyes had moved to the side of the skull, and the nostrils were located further up the skull. The ear showed specialisations for hearing in water.

Legs became shorter

40 mya Dorudon

Dorudon was fully aquatic. Its adaptations for swimming included a long, streamlined body, a broad powerful muscular tail, the development of flippers and webbing. It had very small hind limbs (not attached to the spine) which would no longer bear weight on land.

Hind limbs became detached from spine

Redrawn from de Muizon Nature 2001 413 pp259-260

The hind limbs became fully internal and vestigial. Studies of modern whales show that limb development begins, but is arrested at the limb bud stage. The nostrils became modified as blowholes. This recent ancestor to modern whales diverged into two groups (toothed and baleen) about 36 million years ago. Baleen whales have teeth in their early fetal stage, but lose them before birth.

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Hind limbs are internal and vestigial (have lost their original function).

Balaena (recent whale ancestor)

2. Briefly describe the adaptations of whales for swimming that evolved over time:

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1. Why does the whale fossil record provide a good example of the evolutionary process?

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143 The Evolution of Horses

Key Idea: The evolution of the horse is one of the most robust examples of evolution documented in the fossil record. The evolution of the horse from the ancestral Hyracotherium to modern Equus is well documented in the fossil record. The rich fossil record, which includes numerous transitional fossils, has enabled scientists to develop a robust model

The cooler climates that prevailed in the Miocene (23 -5 mya) brought about a reduction in forested areas with grasslands becoming more abundant. The change in vegetation resulted in the equids developing more durable teeth to cope with the harsher diet. Over time the equid molar became longer and squarer with a hard cement-like covering to enable them to grind the grasses which became their primary diet.

Equus

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of horse phylogeny. It is a complex tree-like lineage with many divergences (below), and a diverse array of often coexisting species. The environmental transition from forest to grasslands drove many of the changes observed in the fossil record. These include reduction in toe number, increased size of cheek teeth, and increasing body size.

5

10

1.6 m

Enamel

15

Dentine

Merychippus

Cement

Millions of years ago

20

Hyracotherium molar

25

1.25 m

30

Mesohippus

35

0.6 m

40

45

50

Hyracotherium (Eohippus)

55

0.4 m

Equus molar

(molars)taller became largertoand more TheCheek equidsteeth also became and faster enable with moretheir cement-like covering. them todurable view and escape predators. This is evident in their overall increase in size and the elongation of their limbs. The reduction in the number of toes from four to one (left) also enabled them to run faster and more efficiently. Most equid evolution took place in North America, although now extinct species did migrate to other regions at various times. During the late Pliocene (2.6 mya) Equus spread into the Old World and diversified into several species including the modern zebras of Africa (right) and the true horse, Equus ferus caballus. Ironically, the horse became extinct in the Americas about 11,000 years ago, and was reintroduced in the 16th century by Spanish explorers.

(a) Change in tooth structure:

(b) Limb length:

(c) Reduction in number of toes:

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1. Explain how the environmental change from forest to grassland influenced the following aspects of equid evolution:

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2. In which way does the equid fossil record provide a good example of the evolutionary process?

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144 Continental Drift and Evolution

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Key Idea: Fossils of related organisms found on separated continents can be explained by continental drift. Continental drift (the movement of the Earth's continents relative to each other) is a measurable phenomenon; it has happened throughout Earth's history and continues today. Movements of up to 2-11 cm a year have been recorded between continents using laser technology. The movements of

the Earth’s 12 major crustal plates are driven by a geological process known as plate tectonics. Some continents are drifting apart while others are moving together. Many lines of evidence show that the modern continents were once joined together as ‘supercontinents’. One supercontinent, Gondwana, was made up of the southern continents some 200 million years ago.

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Greenland

Europe

North America

Asia

Africa

India

South America

New Guinea

New Caledonia

Madagascar

Australia

New Zealand

Antarctica

Key

Glossopteris is a hardy plant that grew adjacent to the glacial ice sheets of Gondwana some 350-230 million years ago

Direction of ice sheet movement 350-230 million years ago

Lystrosaurus is a primitive therapsid (mammal-like) reptile 1 m long, that was widely distributed throughout the southern continents about 240 million years ago.

Old Precambrian rocks (older than 650 mya)

Geomagnetic pole direction 150 million years ago

Precambrian basement rocks (650-570 mya)

Distribution of Lystrosaurus

Distribution of Glossopteris

Early Palaeozoic Late PalaeozoicLate Mesozoic folding (570Early Mesozoic folding folding (160350 mya) (350-160 mya) 70 mya)

1. Name the modern landmasses (continents and large islands) that made up the supercontinent of Gondwana:

Once you have positioned the modern continents into the pattern of the supercontinent, mark on the diagram: (a) The likely position of the South Pole 350-230 million years ago (as indicated by the movement of the ice sheets). (b) The likely position of the geomagnetic South Pole 150 million years ago (as indicated by ancient geomagnetism). State what general deduction you can make about the position of the polar regions with respect to land masses:

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2. Cut out the southern continents opposite and arrange them to recreate the supercontinent of Gondwana. Take care to cut the shapes out close to the coastlines. When arranging them into the space showing the outline of Gondwana on page 207, take into account the following information: (a) The location of ancient rocks and periods of mountain folding during different geological ages. (b) The direction of ancient ice sheet movements. (c) The geomagnetic orientation of old rocks (the way that magnetic crystals are lined up in ancient rock gives an indication of the direction the magnetic pole was at the time the rock was formed). (d) The distribution of fossils of ancient species such as Lystrosaurus and Glossopteris.

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New Zealand

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North America

South America

Greenland

Antarctica

Africa

Europe

Madagascar

India

Asia

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Australia

New Guinea

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Cut out the continental land masses that make up the supercontinent of Gondwana and stick them into the space on the previous page


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This page has been deliberately left blank

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Gondwana supercontinent coastline about 250-150 million years ago

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5. Fossils of Lystrosaurus are known from Antarctica, South Africa, India and Western China. With the modern continents in their present position, Lystrosaurus could have walked across dry land to get to China, Africa and India. It was not possible for it to walk to Antarctica, however. Explain the distribution of this ancient species in terms of continental drift:

6. The southern beech (Nothofagus) is found only in the southern hemisphere, in such places as New Caledonia, New Guinea, eastern Australia (including Tasmania), New Zealand, and southern South America. Fossils of southern beech trees have also been found in Antarctica. They have never been distributed in South Africa or India. The seeds of the southern beech trees are not readily dispersed by the wind and are rapidly killed by exposure to salt water.

(a) Suggest a reason why Nothofagus is not found in Africa or India:

(b) Use a coloured pen to indicate the distribution of Nothofagus on the current world map (on the previous page) and on your completed map of Gondwana above.

(c) State how the arrangement of the continents into Gondwana explains this distribution pattern:

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8. Explain how continental drift provides evidence to support evolutionary theory:

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7. The Atlantic Ocean is currently opening up at the rate of 2 cm per year. At this rate in the past, calculate how long it would have taken to reach its current extent, with the distance from Africa to South America being 2300 km (assume the rate of spreading has been constant):


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145 Speciation and Dispersal of the Camelidae  When very closely related animals were present at

the same time in widely separated parts of the world, it is highly probable that there was no barrier to their movement in between the localities in the past.  The most effective barrier to movement of land animals (especially mammals) was a sea between continents.  The scattered distribution of living species can be explained by migration out of their original range or extinction in regions between the current populations.

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Key Idea: The current fragmented distribution of the camel family can be explained by natural phenomena such as migration, plate tectonics, and changes in sea level. The camel family, Camelidae, consists of six modern-day species on three continents: Asia, Africa and South America. Geophysical phenomena such as plate tectonics and glacial cycles have determined their present, fragmented distribution. Three principles about the dispersal and distribution of land animals are:

Origin and dispersal of the camel family Recent distribution

Tertiary distribution

Arabian camel from North Africa and the Middle East

Vicuña Vicugna vicugna

Ancestor of camel family originated in North America during the tertiary period about 40 million years ago.

South America

Four llama species, including the domesticated llama and alpaca, as well as the wild guanaco and vicuña, exist in the mountainous regions of South America.

North America

Arabian camel Camelus dromedarius

Africa

Asia

Formation of a land bridge across the Bering Strait during a glacial allows passage into Asia by about 1 million years ago.

Bactrian camels in the Gobi Desert region of central Asia.

Bactrian camel Camelus bactrianus

Arabian camels were introduced into Australia from the Middle East in the 1850s. Thousands roam wild throughout Australia’s sandy deserts.

Guanaco Lama guanicoe

Australia

Llama Lama glama

1. Arabian camels are found wild in the Australian outback.

(a) How did they get there?

(b) Why were they absent during prehistoric times?

2. The camel family originated in North America. Suggest why there are no members of the family in North America now:

(b) Explain why it is scattered (discontinuous):

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4. (a) Describe the present distribution of the camel family:

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3. Suggest how early camels managed to get to Asia from North America:

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146 Biogeographical Evidence

Key Idea: Biogeography provides evidence for how the evolution of species is influenced by isolation, continental influences, geological processes, and dispersal ability. Biogeography is the study of the geographical distribution

of species. As described in this activity, island biogeography demonstrates that species on a given island may more closely resemble species on a nearby mainland, rather than species on a distant island with a similar environment.

Galápagos and Cape Verde islands

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Biologists did not fully appreciate the uniqueness and diversity of tropical island biota until explorers began to bring back samples of flora and fauna from their expeditions in the 19th century. The Galápagos Islands, the oldest of which arose 3-4 million years ago, had species similar to but distinct from those on the South American mainland. Similarly, in the Cape Verde Islands, species had close relatives on the West Africa mainland. This suggested to biologists that ancestral forms found their way from the mainland to the islands where they then underwent evolutionary changes.

Galapagos Is

South America

900 km

Cape Verde Is

Western Africa

450 km

Pacific Ocean

South America

Atlantic Ocean

South Atlantic Ocean

4500 km

Africa

3000 km

Tristan da Cunha

Tristan da Cunha species

Tristan da Cunha

The island of Tristan da Cunha in the South Atlantic Ocean is a great distance from any other land mass. Even though it is closer to Africa, there are more species closely related to South American species found there (see table on right). This is probably due to the predominant westerly trade winds from the direction of South America. The flowering plants of universal origin are found in both Africa and South America and could have been introduced from either land mass.

South American origin 7 Flowering plants 5 Ferns 30 Liverworts African origin 2 Flowering plants 2 Ferns 5 Liverworts

Universal origin 19 Flowering plants

1. The Galápagos and the Cape Verde Islands are both tropical islands close to the equator, yet their biotas are quite different. Explain why this is the case:

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2. Explain why the majority of the plant species found on Tristan da Cunha originated from South America, despite its greater distance from the island:

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3. Using one of more examples, describe how biogeography provides support for the theory of evolution:

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147 Oceanic Island Colonisers

Key Idea: The biodiversity of oceanic islands often depends on the distance to the mainland and the ability of plants and animals to survive dispersal. Oceanic islands have a unique biota because only certain groups of plants and animals tend to colonise them. The animals that successfully colonise oceanic islands have to

Land mammals: Few non-flying mammals colonise islands, unless these are very close to the mainland. Mammals have a higher metabolism, need more food and water than reptiles, and cannot sustain themselves on long sea journeys.

be marine in habit or able to survive long periods at sea or in the air. Plants also have limited capacity to reach distant islands. Only some have seeds and fruit that are salt water tolerant. Many plants are transferred to the islands by wind or migrating birds. The biota of the Galápagos islands provide a good example of the result of such a colonisation event.

Small birds, bats, and insects: These animals are blown to islands by accident. They must adapt to life there or perish.

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Blown by strong winds

Plants: Plants have limited capacity to reach distant islands. Only some have fruits and seeds that are salt tolerant. Many plants are transported to islands by wind or birds.

Reptiles: Reptiles probably reach distant islands by floating in driftwood or on mats of floating vegetation. A low metabolic rate enables them to survive the long periods without food and water.

Active flight

Oceanic island

Rafting on drifting vegetation

Amphibians: Cannot live away from fresh water. They seldom reach offshore islands unless that island is a continental remnant.

Planktonic crustacean larvae

Sea mammals: Seals and sea lions have little difficulty in reaching islands, but they return to the sea after the breeding season and do not colonise the interior. Crustaceans: Larval stages drift to islands. Crabs often evolve novel forms on islands. Many are restricted to shoreline areas. Some crabs, such as coconut crabs, have adapted to an island niche.

Mfield cc3.0

Elizabeth Bay

Deep ocean

Swimming

Seabirds: Seabirds fly to and from islands with relative ease. They may become adapted to life on land, as the flightless cormorant has done in the Galápagos. Others, like the frigate bird, may treat the island as a stopping place.

Mauritius, Farquhar, and Diego Rodriguez. These were almost completely exterminated by early Western sailors, although a small population remains on the island of Aldabra. Another feature of oceanic islands is the adaptive radiation of colonising species into different specialist forms. The three species of Galápagos iguana almost certainly arose,

through speciation, from a hardy traveller from the South American mainland. The marine iguana (above) feeds on shoreline seaweeds and is an adept swimmer. The two species of land iguana (not pictured) feed on cacti, which are numerous. One of these (the pink iguana) was identified as a separate species only in 2009.

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The flightless cormorant (above) is one of a number of bird species that lost the power of flight after becoming an island resident. Giant tortoises, such as the 11 subspecies remaining on the Galápagos today (centre) were, until relatively recently, characteristic of many islands in the Indian Ocean including the Seychelles archipelago, Reunion,

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1. Describe one feature typical of an oceanic island coloniser and explain its significance:

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Tiaris bicolor (mainland relative)

Warbler finches

Cocos Island finch

Sharp beaked ground finch

Vegetarian finch

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Tree finches

Cactus ground finches

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A relatively recent (2005) revision of the phylogeny of the finches is shown right. It places the sharp beaked finch more distant from the other ground finches with which it was previously grouped (all are currently Geospiza). A 2015 revision based on whole genome sequencing has split the sharp beaked finches still further into three distinct groups and has suggested a reclassification. The diagram below shows the islands on which the birds are found and the age of the islands in millions of years (in brackets).

Other ground finches

211 The Galápagos Islands, off the west coast of Ecuador, consist of 16 main islands and six smaller islands. They are home to 14 species of finches, each of which has evolved from a single species of grassquit, which arrived from Ecuador. A fifteenth species inhabits Cocos Island. After colonising the islands, the grassquits diversified in response to the availability of unexploited feeding niches. This adaptive radiation is most evident in the beaks of the different species, which are adapted for different purposes, including crushing seeds, pecking wood, or probing cactus flowers.

The five species of tree finches are largely arboreal (tree dwelling). Their sharp beaks are well suited to grasping insects which form the most of their diet. One species has demonstrated tool use in extracting insects.

Darwin (~0.7)

Wolf (1.0)

The Cocos Island finch is the only one of Darwin's finches to be found outside the Galápagos. DNA analysis shows it is related to the warbler finches. The ancestral finch therefore colonised the Galápagos before colonising Cocos Island.

On Wolf Island, the sharp beaked ground finch has become a specialist blood feeder. Recent genomic analysis has suggested a classification of this species.

Cocos Is

800 km

Galápagos Is. 900 km

Pinta (0.8)

N

South America

Pacific Ocean

Marchena (0.7)

Genovesa (0.3)

Equator

The cactus ground finches (left) have evolved probing beaks to extract seeds and insects from cacti.

Santiago Is. (0.8)

Fernandina (0.7)

Santa Cruz (1.1)

Santa Fé (2.7)

Isabela Is (0.5)

San Cristóbal (2.4)

Ground finches, genus Geospiza (left) have crushing type beaks for seed eating. Three species differ mainly in body size and in the size of their beaks. The other three species have longer beaks and supplement their seed diet with cactus flowers and pulp (the cactus finches) or the eggs and blood of other birds and reptile ticks (the sharp-beaked ground finch).

Española Is. (3.2)

Floreana (~1.0)

The beak of the warbler finch is the thinnest of the Galápagos finches. It uses it to spear insects and probe flowers for nectar. It is also the most widespread species, found throughout the archipelago.

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3. Is there are pattern to how the islands were colonised by the birds? Explain:

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2. How do we know the finches colonised Cocos Island after colonising the Galápagos Islands?


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148 Homologous Structures

Key Idea: Homologous structures (homologies) are structural similarities present as a result of common ancestry. The common structural components have been adapted to different purposes in different taxa. The bones of the forelimb of air-breathing vertebrates are composed of similar bones arranged in a comparable pattern. This is indicative of common ancestry. The early

land vertebrates were amphibians with a pentadactyl limb structure (a limb with five fingers or toes). All vertebrates that descended from these early amphibians have limbs with this same basic pentadactyl pattern. They also illustrate the phenomenon known as adaptive radiation, since the basic limb plan has been adapted to meet the requirements of different niches.

Generalised pentadactyl limb

Specialisations of pentadactyl limbs

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The forelimbs and hind limbs have the same arrangement of bones but they have different names. In many cases, the basic limb plan has been adapted (e.g. by loss or fusion of bones) to meet the requirements of different niches (e.g. during adaptive radiation of the mammals). Forelimb

Hind limb

Mole forelimb

Bird wing

Humerus (upper arm)

Femur (thigh)

Dog front leg

Fibula Tibia

Bat wing

Radius Ulna

Carpals (wrist)

Tarsals (ankle)

Metacarpals (palm)

Metatarsals (sole)

Seal flipper

Phalanges (fingers)

Phalanges (toes)

Human arm

1. Briefly describe the purpose of the major anatomical change that has taken place in each of the limb examples above:

(a) Bird wing: Highly modified for flight. Forelimb is shaped for aerodynamic lift and feather attachment.

(b) Human arm:

(c) Seal flipper:

(d) Dog front leg:

(e) Mole forelimb:

(f) Bat wing:

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2. Explain how homology in the pentadactyl limb is evidence for adaptive radiation:

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3. Homology in the behaviour of animals (for example, sharing similar courtship or nesting rituals) is sometimes used to indicate the degree of relatedness between groups. How could behaviour be used in this way:

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149 Developmental Evidence for Evolution

Key Idea: Similarities in the development of embryos, including the genetic control of that development, provides strong evidence for evolution. Developmental biology studies the processes by which organisms grow and develop. In the past, it was restricted to

the appearance of a growing embryo. Today, developmental biology focuses on the genetic control of development and its role in producing the large differences we see in the adult appearance of different species. Differences in gene expression during development are behind these differences. 60

Developmental biology

50

Human

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During development, vertebrate embryos pass through the same stages, in the same sequence, regardless of the total time period of development. This similarity is strong evidence of their shared ancestry. The stage of embryonic development is identified using a standardised system based on the development of structures, not by size or the number of days of development. The Carnegie stages (right) cover the first 60 days of development.

Like humans, mice have digits that become fully separated by interdigital apoptosis during development. In bat forelimbs, this controlled destruction of the tissue between the forelimb digits is inhibited. The developmental program is the result of different patterns of expression of the same genes in the two types of embryos.

Mouse

Chicken

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Days

Carnegie stage of embryonic development Stage 14 Limb buds

IV

III

Stage 17 Digits form

Stage 23 Digits separate

Apoptosis suppressed by expression of an inhibitory growth factor V

II

I

Bat wings are highly specialised structures with unique features, such as elongated wrist and fingers (I-V) and membranous wing surfaces. The forelimb structures of bats and mice are homologous, but how the limb looks and works is quite different.

20

0

As we have seen, homology (e.g. in limb structure) is evidence of shared ancestry. How do these homologous structures become so different in appearance? The answer lies in the way the same genes are regulated during development.

Rhesus monkey

30

10

Limb homology and the control of development

All vertebrate limbs form as buds at the same stage of development. At first, the limbs resemble paddles, but apoptosis (programmed cell death) of the tissue between the developing bones separates the digits to form fingers and toes.

40

Increased expression of a bone growth regulator elongates digits.

IV III

I II

V

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1. Describe a feature of vertebrate embryonic development that supports evolution from a common ancestor:

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2. Explain how different specialised limb structures can arise from a basic pentadactyl structure:

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150 Patterns of Evolution is called an adaptive radiation. The example below (right) describes the radiation of the mammals after the extinction of the dinosaurs made new niches available. Note that the evolution of species may not necessarily involve branching. A species may accumulate genetic changes that, over time, result in the emergence of what can be recognised as a different species. This is known as phyletic gradualism (also called sequential evolution or anagenesis).

Mammalian adaptive radiation

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Key Idea: Populations moving into a new environment may diverge from their common ancestor and form new species. The diversification of an ancestral group into two or more species in different habitats is called divergent evolution. This process is shown below, where two species have diverged from a common ancestor. Divergence is common in evolution. When divergent evolution involves the formation of a large number of species to occupy different niches, this Species P

Species H

Changes in the genetic make-up of the two species

Terrestrial predator niche

Marine predator niche

Arboreal herbivore niche

Underground herbivore niche

Freshwater predator niche

Divergent evolution or cladogenesis: the lineage splits

Species B

Common ancestor

In phyletic gradualism (anagenesis), genetic changes accumulate to form a new species.

Species W

Species D

Extinction

Stasis: very little genetic change; species remains relatively unchanged

Time

Speciation by budding

Browsing/ grazing niche

Flying predator/ frugivore niche

Megazostrodon: one of the first mammals

The earliest true mammals evolved about 195 million years ago, long before they underwent their major adaptive radiation some 65-50 million years ago. These ancestors to the modern forms were very small (12 cm), many were nocturnal and fed on insects and other invertebrates. Megazostrodon (above) is a typical example. This shrew-like animal is known from fossil remains in South Africa and first appeared in the Early Jurassic period (about 195 million years ago). It was climatic change as well as the extinction of the dinosaurs (and their related forms) that suddenly left many niches vacant for exploitation by such adaptable 'generalists'. All modern mammal orders developed very quickly and early.

Species D

Common ancestor

1. In the hypothetical example of divergent evolution illustrated above, left:

(a) Describe the type of evolution that produced species B from species D:

(b) Describe the type of evolution that produced species P and H from species B:

(c) List all species that evolved from: Common ancestor D:

(d) Explain why species B, P, and H all possess a physical trait not found in species D or W:

Common ancestor B:

2. (a) Explain the distinction between divergence and adaptive radiation:

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(b) What are the differences between sequential evolution and divergent evolution?

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151 The Rate of Evolutionary Change

Key Idea: A new species may form gradually over a long period of time, or relatively suddenly. Two main models have been proposed for the rate at which evolution occurs: gradualism and punctuated equilibrium. It is likely that both mechanisms operate at different times and in different situations. Interpretations of the fossil record vary Parent species

New species

New species

Parent species

PR E O V N IE LY W

New species

depending on the time scales involved. During its formative millennia, a species may have accumulated changes gradually (e.g. over 50,000 years). If that species survives for 5 million years, the evolution of its defining characteristics would have been compressed into just 1% of its evolutionary history. In the fossil record, the species would appear quite suddenly.

New species bud off from the parent species and undergo rapid change, followed by a long period of stability.

Each species undergoes gradual changes in its genetic makeup and phenotype.

A typical pattern

A typical pattern

New species diverges from the parent species.

Punctuated equilibrium

Phyletic gradualism

There is abundant evidence in the fossil record that, instead of gradual change, species stayed much the same for long periods of time (called stasis). These periods were punctuated by short bursts of evolution which produce new species quite rapidly. According to the punctuated equilibrium theory, most of a species’ existence is spent in stasis and little time is spent in active evolutionary change. The stimulus for evolution occurs when some crucial aspect of the environment changes.

Phyletic gradualism assumes that populations slowly diverge by accumulating adaptive characteristics in response to different selective pressures. If species evolve by gradualism, there should be transitional forms seen in the fossil record, as is seen with the evolution of the horse. Trilobites, an extinct marine arthropod, are another group of animals that have exhibited gradualism. In a study in 1987 a researcher found that they changed gradually over a three million year period.

1. What rate of environmental change would support the following paces of evolution?

(a) Punctuated equilibrium:

(b) Gradualism:

2. In the fossil record of early human evolution, species tend to appear suddenly, linger for often very extended periods before disappearing suddenly. There are few examples of smooth inter-gradations from one species to the next. Which of the above models best describes the rate of human evolution?

3. Some species show apparently little evolutionary change over long periods of time (hundreds of millions of years). (a) Name two examples of such species:

(b) What term is given to this lack of evolutionary change?

(c) Such species are often called 'living fossils'. Why might they have been described in this way?

(d) Why would such species change so little over evolutionary time?

CL

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LINK

WEB

184 151

KNOW


152 Divergent Evolution in Ratites

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216

of ratites were flying birds living about 80 million years ago. These ancestors also had a primitive palate, but they possessed a keeled breastbone. Flightlessness in itself is not unique to ratites; there are other birds that have lost the power of flight, particularly on remote, predator-free islands. All ratites have powerful legs, and many, such as the emu, can run very quickly.

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Key Idea: The ratites are group of birds descended from a single common ancestor that lost the power of flight very early on in their evolutionary development. Ratites are flightless birds that possess two features that distinguish them from other birds; a flat breastbone (instead of the more usual keeled shape) and a primitive palate (roof to the mouth). Fossil evidence indicates that the ancestors

Elephant bird Two species, extinct, Madagascar

Ostrich Struthio camelus, Africa

Cassowary Three species, Australia & New Guinea.

Emu Dromaius novaehollandiae, Australia.

Rhea Two species, South America.

Kiwi Five species, New Zealand.

*Tinamous from South America were until recently thought to be related to but not part of the ratite group. New evidence suggests they should be included in ratites.

It had long been thought that the geographical distribution of modern day and extinct ratite species could be explained by continental drift. The "rafting hypothesis" suggests that the ancestral ratite population existed when the southern continents of South America, Africa, and Australia (and their major offshore islands) were joined as a single land mass called Gondwana. As the continents moved apart as a result of plate tectonics, the early ratite populations were carried with them. Mitochondrial DNA (mtDNA) evidence now suggests kiwis are most related to the extinct elephant bird from Madagascar and slightly less closely related to emus in Australia. However, the ancestor to the kiwi arrived in New Zealand long after New Zealand separated from the rest of Gondwana. Ancestral kiwi must therefore have flown there. Moas are now thought to be closely related to tinamous (South America), which can fly. Ostriches were thought to be closely related to elephant birds but mtDNA now suggests they diverged from the other ratites early. The conclusions from these new findings suggest that the ratites evolved from flighted birds that flew between continents and independently evolved flightlessness at least six times.

Moa Eleven species (Lambert et al. 2004*), all extinct, New Zealand.

Ratite phylogeny

Moas diverged from tinamous about 55 million years ago.

Moa

Tinamou Elephant bird

N AS OT SR F OO OR M US E

Birds evolved from a theropod dinosaur ancestor about 150 million years ago.

Ratites diverged from other birds between 90 and 70 million years ago.

Kiwi

Emu

Cassowary Rhea

Ostrich

Other birds

LINK

KNOW

163

50 mya

25 mya

CL

75 mya

Present

From Mitchell et al

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1. (a) Describe three physical features distinguishing all ratites (excluding tinamous) from most other birds:

(b) Why should tinamous be included in ratites?

PR E O V N IE LY W

2. Describe two anatomical changes, common to all ratites (excluding tinamous), which have evolved as a result of flightlessness. For each, describe the selection pressures for the anatomical change:

(a) Anatomical change:

Selection pressure:

(b) Anatomical change:

Selection pressure:

3. (a) Name two other flightless birds that are not ratites:

(b) Why are these other flightless species not considered part of the ratite group?

4. Kiwis are ratites that have remained small. They arrived in New Zealand long after the moa. What part might this late arrival have played in kiwi species remaining small?

5. (a) On the phylogenetic tree opposite, circle the branching marking the common ancestor of moa and kiwi. (b) On the phylogenetic tree opposite, circle the branching marking the common ancestor of emus and kiwi. 6. (a) Based on the rafting hypothesis which ratite would you expect to be most closely related to ostriches?

(b) Which ratite group is actually the closest related to the ostrich?

CL

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7. The diversification of ratites may still be explained in part by continental drift. Use the data on the opposite page to suggest a possible sequence of events for the distribution of ratites:


153 Adaptive Radiation in Mammals

10,000 yrs

Marsupials

Rodents

Colugos

Bats

Insectivores

Pinnipoeds

Carnivores

Whales & dolphins

Odd-toed ungulates

Aardvark

Even-toed ungulates

Elephants

Hyraxes

Holocene

Sea-cows

Geologic time scale

Monotremes

Hares, rabbits Pangolins Anteaters, sloths

diagram below shows the divergence of the mammals into major orders, many occupying niches left vacant by the dinosaurs. The vertical extent of each grey shape shows the time span for which a particular order has existed. Those that reach the top of the chart have survived to the present day. The width of a grey shape shows how many species existed at any given time. The dotted lines indicate possible links between the orders for which there is no direct fossil evidence.

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Key Idea: An ancestral mammal group underwent adaptive radiation about 80 mya and diversified into all the mammalian groups seen today. Adaptive radiation is diversification among the descendants of a single ancestral group to occupy different niches. Mammals underwent an extensive adaptive radiation following the extinction of the dinosaurs. Most of the modern mammalian groups became established very early on. The

Primates Elephant shrews

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218

Pleistocene

1.8 my

Pliocene

5 my

Miocene

25 my

Oligocene

37 my

A

Eocene

53 my

Paleocene

A

D

D

65 my

Cretaceous

B

134 my

Jurassic

Early mammal Megazostrodon

200 my

Triassic

C

1. In general terms, describe the adaptive radiation that occurred in mammals:

2. Name the term that you would use to describe the animal groups at point C (above): 3. Explain what occurred at point B (above):

N AS OT SR F OO OR M US E

4. Describe one thing that the animal orders labelled D (above) have in common:

5. Identify the two orders that appear to have been most successful in terms of the number of species produced:

6. Explain what has happened to the mammal orders labelled A in the diagram above:

LINK

KNOW

150

CL

7. Name the geological time period during which there was the most adaptive radiation:

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Rodent biodiversity Rodents make up 40% of mammalian species, making them easily the most successful of the mammalian groups. Fossils with distinctive rodent features first appeared about 66 million years ago. All rodents have upper and lower incisor teeth that grow continuously. They are morphologically quite generalised and highly adaptable, occupying a wide number of habitats from deserts to forest. In some cases, distantly related species have occupied the same type of habitat and niche in widely separated regions, e.g. the kangaroo rat in western North America deserts (beaver-like rodents) and the jerboa in African deserts (mouse-like rodents).

Porcupine-like rodents

Squirrels are found on many continents. Their lifestyles include tree dwelling, ground dwelling, and gliding forms. Like most rodents they are social, with prairie dogs forming large communities called towns.

Capybaras are South American rodents and the largest of all rodents. They occupy habitats from forests to savannahs. Porcupines are found throughout the Old and New Worlds. Their spines make an almost impenetrable defense against predators. The group also includes guinea pigs, which are popular as pets.

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Squirrel-like rodents

Guinea pig

Chipmunk

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Porcupine

Squirrel

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Beaver

Central rock rat

Kangaroo rat

White footed mouse

Mouse-like rodents

Beavers are one of the larger types of rodents. They live near rivers, streams and lakes, chewing through small trees to build dams across streams and lodges to live in. Gophers live in burrows, while kangaroo rats are so well adapted to the desert they virtually never need to drink.

Rats and mice are found in virtually every part of the world thanks to their generalist adaptations and human assisted travel. There are at least 100 species of rats and mice alone. The group also includes voles, lemmings, jerboas and dormice.

9. Describe some of the habitats rodents have occupied:

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8. What anatomical feature do all rodents have?

N AS OT SR F OO OR M US E

Beaver-like rodents


154 Convergent Evolution

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220

Key Idea: Evolution in response to similar selection pressures can result in unrelated species appearing very similar. Convergent evolution (convergence) describes the process by which species from different evolutionary lineages come to resemble each other because they have similar ecological roles, and natural selection has shaped similar adaptations.

It can be difficult to distinguish convergent and parallel evolution, as both produce similarity of form. Generally, similarity arising in closely related lineages (e.g. within marsupial mice) is regarded as parallelism, whereas similarity arising in more distantly related taxa is convergence (e.g. similarities between marsupial and placental mice).

Convergence: same look, different origins

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Not all similarities between species are the result of common ancestry. Selection pressures to solve similar problems in particular environments may result in similarity of form and function in unrelated (or distantly related) species. The evolution of succulence in unrelated plant groups (Euphorbia and the cactus family) is an example of convergence in plants. In the example (right), the selection pressures of the aquatic environment have produced a similar streamlined body shape in unrelated vertebrates. Icthyosaurs, penguins, and dolphins each evolved from terrestrial species that took up an aquatic lifestyle. Their general body form has evolved to become similar to that of the shark, which has always been aquatic. Note that flipper shape in mammals, birds, and reptiles is a result of convergence, but its origin from the pentadactyl limb is an example of homology (common ancestry).

Fish: Shark

Reptile:Icthyosaur (extinct)

Mammal:Dolphin

Bird: Penguin

Analogous structures (homoplasies) have the same function and often the same appearance, but different origins. The example (right) shows the structure of the eye in two unrelated taxa (mammals and cephalopod molluscs). The eye appears similar, but has evolved independently.

The wings of birds and insects are also analogous. The wings have the same function, but the two taxa do not share a common ancestor. Longisquama, a lizard-like creature that lived about 220 mya, also had ‘wings’ that probably allowed gliding between trees. These 'wings’ were highly modified long scales or feathers extending from its back and not a modification of the forearm (as in birds).

Jo Naylor cc2.0

Analogous structures arise through convergent evolution

Mammalian eye

Retina

Octopus eye

Iris

Iris

Lens Cornea

Lens

Retina

Cornea

1. In the example above illustrating convergence in swimming form, describe two ways in which the body form has evolved in response to the particular selection pressures of the aquatic environment: (a)

(b)

2. Describe two of the selection pressures that have influenced the body form of the swimming animals above: (a) (b)

KNOW

154

CL

WEB

N AS OT SR F OO OR M US E

3. When early taxonomists encountered new species in the Pacific region and the Americas, they were keen to assign them to existing taxonomic families based on their apparent similarity to European species. In recent times, many of the new species have been found to be quite unrelated to the European families they were assigned to. Explain why the traditional approach did not reveal the true evolutionary relationships of the new species:

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4. For each of the paired examples, briefly describe the adaptations of body shape, diet and locomotion that appear to be similar in both forms, and the likely selection pressures that are acting on these mammals to produce similar body forms:

221

Convergence between marsupials and placentals

North America

PR E O V N IE LY W

Australia

Marsupial and placental mammals diverged very early in mammalian evolution (about 120 mya), probably in what is now the Americas. Marsupials were widespread throughout the ancient supercontinent of Gondwana as it began to break up through the Cretaceous, but then became isolated on the southern continents, while the placentals diversified in the Americas and elsewhere, displacing the marsupials in most habitats around the world. Australia's isolation from other landmasses in the Eocene meant that the Australian marsupials escaped competition with the placentals and diversified into a wide variety of forms, ecologically equivalent to the North American placental species.

Placental mammals

Marsupial mammals

Wombat

(a)

Adaptations: Rodent-like teeth, eat roots and

Woodchuck

above ground plants, and can excavate burrows. Selection pressures: Diet requires chisel-like

teeth for gnawing. The need to seek safety from predators on open grassland.

Flying phalanger

(b)

Adaptations:

Flying squirrel

Selection pressures:

Marsupial mole

(c)

Adaptations:

Mole

Selection pressures:

Marsupial mouse

(d)

Adaptations:

Mouse

Selection pressures:

Tasmanian wolf (tiger)

(e)

Adaptations:

Wolf

Jack rabbit

(f)

Adaptations:

Selection pressures:

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Long-eared bandicoot

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Selection pressures:


155 Extinction

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222

Either it evolves into a new species (and the ancestral species becomes effectively extinct) or it and its lineage becomes extinct. A mass extinction describes the widespread and rapid (in geologic terms) decrease in life on Earth and involves not only the loss of species, but the loss of entire families (which are made up of many genera and species). Such events are linked to major climate shifts or catastrophic events. There have been five previous mass extinctions. The diagram below shows how the diversity of life has varied over the history of life on Earth and aligns this with major geologic and climatic events.

PR E O V N IE LY W

Key Idea: Extinction is a natural process. Mass extinctions involve the extinction of large numbers of species and genera in a geologically short period of time. Extinction is the death of an entire species, no individuals are left alive. Extinction is an important (and natural) process in evolution and describes the loss of a species forever. Extinction provides opportunities, in the form of vacant niches, for the evolution of new species. More than 98% of species that have ever lived are now extinct, most of these before humans were present. Extinction is the result of a species being unable to adapt to an environmental change.

700

Numbers of families

600 500 400 300 200 100

500

400

300

200

100

0

12 10 8 6 4 2 0 -2 -4 -6 -8

Warm climate

Cool climate

300 200 100 0

-100

50

0 Palaeozoic 500

WEB

KNOW

LINK

155 150

400

N AS OT SR F OO OR M US E

150 100

Mesozoic 300

200

Cenozoic

100

CL

Meterorite impacts (crater radius, km)

Large scale volcanism

Changes in sea level (m)

Changes in temperature (°C)

Millions of years ago (MYA)

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0


Devonian extinction (375-360 MYA). Marine life affected especially brachiopods, trilobites, and reef building organisms.

Permian extinction (252 MYA). Nearly all life on Earth perished. 57% of families and 83% of genera were wiped out. 96% of marine species became extinct.

Conodont Triassic extinction (201 MYA). At least half of the species present became extinct, vacating niches and ushering in the age of the dinosaurs.

RA

Coral

Dinosaur Cretaceous extinction (66 MYA). Marked by the extinction of nearly all dinosaur species (their descendants, the birds, survive).

PR E O V N IE LY W

Ordovician extinction (458-440 MYA). Second largest extinction of marine life: >60% of marine invertebrates died. One of the coldest periods in Earth's history.

Trilobite

USGS

Graptolite

Moussa Direct Ltd.

Wilson44691

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223

1. What is the general cause of extinction?

2. What is a mass extinction?

3. Why is extinction an important natural process?

4. Study the data on the opposite page carefully and answer the following questions: (a) What is the evidence that extinction is a natural process?

(b) Mark the five mass extinctions on the top graph opposite:

(c) Is there reason to believe that there is any one cause for any of these mass extinctions? Explain your answer:

(d) What has happened to the diversity of life soon after each mass extinction?

6. Which extinctions mark the rise and end of the dinosaurs?

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5. Which of the five extinctions appears to be the most severe?

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156 KEY TERMS AND IDEAS: Did You Get It?

1. Test your vocabulary by matching each term to its correct definition, as identified by its preceding letter code. This activity tests your knowledge of the vocabulary used in this and the previous chapter.

adaptive radiation

analogous structure

A feature of different species that are similar in function but not necessarily in structure and in which the similarity is not the result of common ancestry.

B

Anatomical similarities in organisms occurring as a result of their common ancestry.

C

A form of divergent evolution in which there is rapid speciation of one ancestral species to fill many different ecological niches.

PR E O V N IE LY W

chronometric dating

A

common ancestor

convergent evolution divergent evolution extinction fossil

homologous structure

D

The science determining the relative order of past events, without necessarily determining their absolute age.

E

The division of one species, during evolution, into two or more separate species.

F

The complete dying out of a species so that there are no representatives of the species remaining anywhere.

G

The most recent individual from which all organisms in the group are directly descended.

H

The evolutionary history of a group of organisms. Often represented as a 'tree' showing descent of new species from an ancestral one.

I

Determining an approximate age for an archaeological or palaeontological site or artefact based on its physical or chemical properties of the materials.

J

The preserved remains or traces of organisms from the remote past.

K

Evolution in unrelated species occupying similar niches that causes them to arrive at similar structural, physiological and behavioural solutions.

phylogeny

relative dating

2. (a) The two plants pictured right are unrelated. The left image shows a cactus from North America, while the right image shows a Euphorbia from Africa. Both these plants live in deserts. Identify the pattern of evolution shown here:

(b) Describe the environment associated with the adaptations:

3. (a) The diagrams above right show the forelimbs of a whale, bird, and human. Shade the diagram to indicate which bones are homologous. Use the same colour to indicate the equivalent bones in each limb. (b) What does the homology of these bones indicate?

N AS OT SR F OO OR M US E

TEST

Bird

CL

Human

Whale

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Determining relatedness between species

PR E O V N IE LY W

Unit 4 Outcome 1

Key terms

Molecular homology

BMP4

Key knowledge

c

1

Discuss the molecular evidence that reveals similarities between closely related organisms with reference to DNA and protein sequence data, mitochondrial DNA (mtDNA), and genome-wide comparisons (including DNA hybridisation).

157 - 161

c

2

Describe how biochemical variations can be used as a molecular clock to determine the probable dates of divergence from a common ancestor. How are molecular clocks calibrated and what are their benefits and disadvantages?

160 161

The use of phylogenetic trees

Activity number

evo-devo

master gene

molecular clock mtDNA

phylogenetic tree phylogeny

Key knowledge

162

c

3

Explain how phylogenetic trees are used to show relatedness between species. What are phylogenetic trees based on and how are they constructed?

c

4

Recognise a cladogram as a phylogenetic tree produced using cladistic analysis (analysis of shared derived characteristics). As extension, construct a cladogram to show the evolutionary history of a group of related organisms.

163 164

c

5

Explain how phylogenetic trees can change in the light of new information or improved technology. Use an example to illustrate your point.

163 168 184

Putneymark cc 2.0

Master genes and evolution of new forms Key knowledge

Activity number

c

6

Recall how evolutionary developmental biology (evo-devo) is helping us to understand how changes to the way in which genes are regulated can produce major changes to phenotype.

149

c

7

Outline the role of master genes in regulating animal development. Explain how small changes to the expression of these genes (e.g. the Hox, Pax, and BMP genes) can result in the evolution of novel phenotypes.

157 165 166

c

8

With reference to the gene BMP4, explain how differences in gene expression can produce differences in morphology, e.g. beak differences in Galápagos finches and different jaw morphologies in African cichlids.

166

N AS OT SR F OO OR M US E

DNA hybridisation

CL

common ancestor

Activity number


157 Determining Relatedness Using Proteins

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226

Key Idea: Proteins are the product of gene expression, so an analysis of the differences between the same protein in different taxa gives an indication of species relatedness. Traditionally, phylogenies were based largely on anatomical traits, and biologists attempted to determine the relationships between taxa based on similarity or by tracing the appearance of key characteristics. With the advent of new molecular techniques, homologies (similarities arising from shared

ancestry) could be studied at the molecular level as well and the results compared to phylogenies established using other methods. Protein sequencing provides an excellent tool for establishing homologies. A protein has a specific number of amino acids arranged in a specific order. Any differences in the sequence reflect changes in the DNA sequence. Commonly studied proteins include blood proteins, such as haemoglobin, and the respiratory protein cytochrome c.

PR E O V N IE LY W

Haemoglobin homology Human – chimpanzee 0

Chicken 45

Horse 25

Gibbon 2

Rhesus monkey 8

Gorilla 1

Dog 15

Mouse 27

Frog 67

Kangaroo 38

Increasing difference in amino acid sequence

Primates

Placental mammals

Marsupial

Non-mammalian vertebrates

Haemoglobin is the oxygen-transporting blood protein found in most vertebrates. The beta chain haemoglobin sequences from different organisms can be compared to determine evolutionary relationships. As genetic relatedness decreases, the number of amino acid differences between the haemoglobin beta chains of different vertebrates increases (above). For example, there are no amino acid differences between humans and chimpanzees, indicating they recently shared a common ancestor. Humans and frogs have 67 amino acid differences, indicating they had a common ancestor a very long time ago.

Highly conserved proteins

The Pax-6 protein provides evidence for evolution

Some proteins are common in many different species. These proteins are called highly conserved proteins, meaning they change (mutate) very little over time. This is because they have critical roles in the organism (e.g. in cellular respiration) and mutations are likely to be lethal.

►► The Pax-6 gene belongs to a family of master genes that regulate the formation of a number of organs, including the eye, during embryonic development. ►► The Pax-6 gene produces the Pax-6 protein, which acts as a transcription factor to control the expression of other genes.

Evidence indicates that highly conserved proteins are homologous and have been derived from a common ancestor. Because they are highly conserved, changes in the amino acid sequence are likely to represent major divergences between groups during the course of evolution.

►► Scientists know the role of Pax-6 in eye development because they created a knockout model in mice where the Pax-6 gene is not expressed. The knockout model is eyeless or has very underdeveloped eyes. ►► The Pax-6 gene is so highly conserved that the gene from one species can be inserted into another species, and still produce a normal eye.

Cytochrome C (left) is a respiratory protein located in the electron transport chain in mitochondria.

Histone protein

KNOW

LINK

DNA LINK

LINK

LINK

157 135 159 160 165

CL

An experiment inserted mouse Pax-6 gene into fly DNA and turned it on in a fly's legs. The fly developed morphologically normal eyes on its legs!

Histones (right) are a family of proteins that associate with DNA and organise it so that it can fit inside the cell nucleus.

WEB

N AS OT SR F OO OR M US E

Emw

►► This suggests the Pax-6 proteins are homologous, and the gene has been inherited from a common ancestor.

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227 Using immunology to determine phylogeny Precipitate forms

Human

Gorilla

Baboon

Lemur

Rat

Decreasing recognition of the antibodies against human blood proteins

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The immune system of one species will recognise the blood proteins of another species as foreign and form antibodies against them. This property can be used to determine the extent of relatedness between species. Blood proteins, such as albumins, are used to prepare antiserum in rabbits, a distantly related species. The antiserum contains antibodies against the test blood proteins (e.g. human) and will react to those proteins in any blood sample they are mixed with. The extent of the reaction indicates how similar the proteins are; the greater the reaction, the more similar the proteins. This principle is illustrated (right) for antiserum produced to human blood and its reaction with the blood of other primates and a rat.

30

Millions of years ago 10 20

0

North American tree frogs

European tree frogs Cricket frog Chorus frogs

Australian tree frog

60

50

40 30 20 10 Immunological distance

0

The relationships among tree frogs have been established by immunological studies based on blood proteins such as immunoglobulins and albumins. The immunological distance is a measure of the number of amino acid substitutions between two groups. This, in turn, has been calibrated to provide a time scale showing when the various related groups diverged.

1. Compare the differences in the haemoglobin sequence of humans, rhesus monkeys, and horses. What do these tell you about the relative relatedness of these organisms?

2. (a) What is a highly conserved protein?

(b) What type of proteins tend to be highly conserved?

(c) Why are the proteins named in (b) highly conserved?

(d) Why are highly conserved proteins good for constructing phylogenies?

(b) What evidence is there that the Pax-6 protein is highly conserved?

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N AS OT SR F OO OR M US E

3. (a) Describe the role of the Pax-6 gene:


158 Determining Relatedness by DNA Hybridisation

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228

Key Idea: DNA hybridisation compares DNA similarity between species and can be used to measure relatedness. DNA hybridisation is a technique used to quantify the DNA similarity between species. More closely related species have fewer genetic differences than more distantly related species. The method provides information only about how much of

DNA hybridisation technique

1. How can DNA hybridisation give a measure of genetic relatedness between species?

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1. DNA from the two species to be compared is extracted, purified and cut into short fragments.

the DNA is the same and cannot provide specific information about what the similarities or differences are. Although it has largely been replaced by DNA sequence analysis, DNA hybridisation is still used in microbial studies and has been used to determine the date of human divergence from apes, which has been estimated at 10 and 5 million years ago.

2. The mixture is heated so the DNA separates. The DNA from the two species is mixed together. 3. As it cools, bonds form between compatible nucleotides. Hybrid double-stranded DNA forms.

4. If species share low similarity, the hybrid DNA will have few bonds (and the strands will be weakly held together). The number of bonds (and therefore the strength of the hybrid DNA) increases with increasing similarity.

5. The similarity is measured by heating the hybrid DNA to force it to form single strands. The greater the similarity, the more heat that is required to break the hybrid DNA apart.

2. Why do the double strands of DNA break when they are heated?

3. What is responsible for the hybridisation between the DNA strands?

Human DNA

Chimpanzee DNA

4. The graph below shows the results of a DNA hybridisation between humans and other primates.

Fragments of double stranded DNA held together by hydrogen bonds.

Similarity of human DNA to that of other primates

0

DNA similarity (%) 40 60

20

80

Human

Cool the samples

94.7%

Rhesus monkey

91.1%

Vervet monkey

90.5%

Capuchin monkey

84.2%

58.0%

(a) Which primate is most closely related to humans?

(b) Which primate is most distantly related to humans?

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These bases do not match LINK

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5. Hybrid DNA from species A and B comes apart at a lower temperature that of species A and C. Which species is A most closely related to? These bases match

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97.6%

Gibbon

Galago

Hybrid DNA

100%

Chimpanzee

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Single-stranded DNA from the two species is mixed

Primate species

Heat the DNA samples. Heat disrupts the hydrogen bonding so the strands separate.

100

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159 Genomic Comparisons and Relatedness

Key Idea: Comparing nucleotide sequences in DNA provides detailed information about relatedness between organisms. DNA sequencing provides the precise order of nucleotides in a DNA molecule. This information, which can now be analysed using sophisticated computing, allows researchers to compare sequences between species in much more detail

than is possible with DNA hybridisation. Not only can areas of difference be identified, but the variation between the nucleotides at a certain position can be determined. This information allows researchers to more accurately determine the relatedness between species, even between those with very minor differences.

Comparing DNA sequences

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Improved DNA sequencing techniques and powerful computing software have allowed researchers to accurately and quickly sequence and compare entire genomes (all an organism's genetic material) within and between species.

Once DNA sequences have been determined, they are aligned and compared to see where the differences occur (right). DNA sequencing generates large volumes of data and the rise in computing power has been central to modern sequence analyses. The technological advances have been behind the new field of bioinformatics, which uses computer science, statistics, mathematics, and engineering to analyse and interpret biological data.

DNA: Species 2

Species 1 Species 2

What type of sequences are compared?

Highly conserved sequences are often used for comparative genomic analysis because they are found in many organisms. The changes (mutations) of the sequences over time can be used to determine evolutionary relationships. As with other forms of molecular analysis, species with fewer nucleotide differences are more closely related than those with many.

Whole genome analysis has been important in classifying the primates. Historical views attributed special status to humans which often confused primate classification schemes. DNA evidence provides impartial quantitative evidence and modern classification schemes have been based on this data.

Based on DNA evidence, chimpanzees are more closely related to humans than they are to gorillas and there is no taxon called "great apes".

1. (a) What advantages does DNA sequence comparison have over DNA hybridisation?

(b) How is this an advantage in determining evolutionary relationships?

2. Three partial DNA sequences for three different species are presented below.

A T G G C C C C C A A C A T T C G A A A A T C G C A C C C C C T GC T C A A A A T T A T C A A C

Species 2

ATGGCACCTAACATCCCCAACTCCCACCGT GTACTCAAAATCATCAAG

Species 3

ATGGCACCCAATATCCGCAAATCACACCCCCTGTTAAAAACAATCAAC

Based on the number of differences in the DNA sequences:

(a) Identify the two species most closely related: (b) Identify the two species that are least closely related: Š2020 BIOZONE International ISBN: 978-1-98-856647-4 Photocopying Prohibited

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Species 1

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160 The Molecular Clock Theory

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Key Idea: The molecular clock hypothesis proposes that mutations occur at a steady rate and that changes in DNA sequences between species can determine phylogeny. The molecular clock hypothesis states that mutations occur at a relatively constant rate for any given gene. The genetic difference between any two species can indicate when two Time 0

+ 25 million years mutation 1 occurred

+ 50 million years mutation 2 occurred

CAATTGATCG

CAATCGATCG

(A)

CAATTTATCT

CAATTTATTT

(B)

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In a theoretical example, the DNA sequence for a gene in two species (A & B, right) alive today differs by four bases. The mutation rate for the gene is approximately one base per 25 million years. Based on this rate, it can be determined that the common ancestor for these two species lived 50 mya.

species last shared a common ancestor and can be used to construct a phylogenetic tree. The molecular clock for each species, and each protein, may run at different rates, so molecular clock data is calibrated with other evidence (e.g. morphological) to confirm phylogeny. Molecular clock calculations are carried out on DNA or amino acid sequences.

Common ancestor CAATTTATCG

Cytochrome c and the molecular clock theory

Human

1

2

3

4

5

6

7

8

9

Gly

Asp

Val

Glu

Lys

Gly

Lys

Lys

Ile

10

Phe Ile

Pig

Chicken

Ile

Dogfish

Drosophila

<<

Wheat

<<

Asn

Pro

Asp

Yeast

<<

Ser

Ala

Lys

Ala

Thr

12

13

14

15

16

17

18

19

20

21

22

Met

Lys

Cys

Ser

Gln

Cys

His

Thr

Val

Glu

Lys

Val

Gln

Ala

Val

Val

Gln

Ala

Val

Val

Gln

Ala

Asn

Leu

Val

Gln

Arg

Ala

Ala

Lys

Thr

Arg

Ala

Lys

Thr

Arg

Glu

Ala Ala

11

Leu

Asp

Ala

Leu

This table shows the N-terminal 22 amino acid residues of human cytochrome c, with corresponding sequences from other organisms aligned beneath. Sequences are aligned to give the most position matches. A shaded square indicates no change. In every case, the cytochrome's heme group is attached to the Cys-14 and Cys-17. In Drosophila, wheat, and yeast, arrows indicate that several amino acids precede the sequence shown. Human Monkey Dog Horse Donkey Pig Kangaroo Rabbit Pigeon Duck Chicken Turtle Rattlesnake Tuna Screwworm fly Samia cynthis (moth) Neurospora crassa (mold) Saccharomyces (baker’s yeast) Candida krusei (yeast)

The sequence homology of cytochrome c (right), a respiratory protein, has been used to construct a phylogenetic tree for some species. Overall, the phylogeny aligns well to other evolutionary data, although the tree indicates that primates branched off before the marsupials diverged from other placental mammals, which is incorrect based on other evidence. Highly conserved proteins, such as cytochrome c, change very little over time and between species because they carry out important roles and if they changed too much they may no longer function properly.

Ancestral organism

25

20

15

10

Average amino acid substitutions

1. Describe a limitation of using molecular clocks to establish phylogeny:

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2. For cytochrome c, suggest why amino acids 14 and 17 are unchanged in all the organisms shown in the table:

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161 Using Mitochondrial DNA

Key Idea: Mitochondrial DNA (mtDNA) can be used to determine relationships between closely related species. Mitochondrial DNA (mtDNA) is a single circular piece of DNA found in the mitochondria of eukaryotic organisms. mtDNA mutates at a much higher rate than nuclear DNA and it is inherited, without the usual genetic recombination*, only

Identifying species using mtDNA Transatlantic mangrove oysters of the genus Crassostrea are important commercial oysters on the Atlantic shores of South America and Africa. mtDNA studies have found that the division of this genus into distinct South American and African species (the African C. gasar and the South American C. rhizophorae) may be incorrect.

PR E O V N IE LY W

The length of the mtDNA can vary between eukaryotes. The human mitochondrial genome contains about 16 kilobases. It encodes 37 genes, 22 of them being for mitochondrial tRNA. Genes that are commonly compared are the 16S rRNA gene and the control region of the mtDNA.

from the mother (except in extremely rare occurrences). Thus mtDNA mutations are passed on 100% of the time from mother to all her offspring. These two features make mtDNA useful for determining relationships between closely related species or individuals within a species, and for following maternal lineages through time.

Control region

16S rRNA

22 tRNA encoding genes

Human mtDNA

A study sequenced a 570 base-pair length of mtDNA from the 16S rRNA gene of 18 individuals from nine locations along the African and South American coastline. The study found two distinct DNA sequences. Bases 1 - 60 are shown below: A TTGATTTTTAGTAGTACCTGCCCAGTGCG-TATTATCTTGTTAACGGCCGCCTT B . . . . . . . . . . . C . . . . . . . . . . . . . . . . . . . A. . . . .AG . C . . C. . . . . . . . . . . . In was presumed that the mangrove oyster Crassostrea gasar was found only on the African side of the Atlantic while Crassostrea rhizophorae was found only in South America. All samples from the African coastline had the same sequence as A above. However the South American sample was found to have both A and B DNA sequences, showing that Crassostrea gasar is also present in South America.

C. gasar on mangrove roots

Mitochondrial Eve

Using mtDNA to trace human ancestry

Because mtDNA is passed through the maternal line without the usual genetic recombination* it can be used to trace maternal lineage. This means that, barring new mutations, the mtDNA of any one person is the same as their direct maternal ancestor back many generations. In humans, this concept has been used to trace the most-recent common mitochondrial ancestor of all humans, a single female from Africa (dubbed Mitochondrial Eve or ME). The ME represents that woman whose mitochondrial DNA (with mutations) exists in all humans alive today. Mutations to the mtDNA provide the molecular clock that allows us to determine how much time has elapsed since the ME lived. The existence of a ME does not mean that no other women have left descendents; we have ancestors who are not via matrilineal descent (e.g. you have nuclear DNA from your both father's and mother's mother, but only mitochondrial DNA from your mother's mother). At some point, these other females must have produced no daughters themselves and so broke the mitochondrial line (right).

Generation 1 2 3

Mutation

4 5 6

*Although mtDNA does recombine, it does so with copies of itself within the same mitochondrion.

1. (a) How does mtDNA differ from nuclear DNA?

(b) Why is mtDNA useful for following maternal lineages?

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2. Why is mtDNA useful for determining relationships between closely related species:

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162 What is a Phylogenetic Tree? of individual species or be a larger group (e.g. an order). Traditionally phylogenetic trees have been constructed based on similarities or differences in appearance, but in more recent times molecular comparisons have been used. Molecular phylogenetics can reveal differences not seen in morphological comparisons, and have resulted in the revision of some morphological phylogenies where organisms could not be separated on appearance. Phylogenetic trees are often constructed based on cladistics (below).

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Key Idea: Phylogenetic trees represent possible evolutionary histories between organisms. They can be constructed based on appearance or molecular data. Phylogenetics is the study of the evolutionary history and the relationships among individuals or groups of organisms. These relationships are often shown diagrammatically as a phylogenetic tree. A phylogenetic tree represents a likely hypothesis of the evolutionary relationships between biological groups or taxa (sing. taxon). A taxon may consist

What do phylogenetic trees look like?

Phylogenetic trees represent possible evolutionary histories, and there are many different ways they can be drawn (right). Depending on how the tree is constructed, some represent evolutionary time through the length of the branches (lines). Phylogenetic trees based on cladistics (cladograms) do not represent evolutionary time.

Determining phylogenetic relationships

►► Increasingly, analyses to determine evolutionary relationships rely on cladistic analyses of character states. Cladistic analysis groups species according to their most recent common ancestor on the basis of shared derived characteristics or synapomorphies. All other characters are ignored.

►► A phylogeny constructed using cladistics thus includes only monophyletic groups, i.e. the common ancestor and all of its descendents. It excludes both paraphyletic and polyphyletic groups (right). It is important to understand these terms when constructing cladograms and to also understand that the terms are relative to whereever you start in the phylogenetic tree (i.e. where the common ancestor is).

►► The cladist restriction to using only synapomorphies creates an unambiguous branching tree. One problem with this approach is that a strictly cladistic classification could theoretically have an impractically large number of taxonomic levels and may be incompatible with a Linnaean system.

D

C B

A

B

C

D

A

Taxon 2 is polyphyletic as it includes organisms with different ancestors. The group "warm-blooded (endothermic) animals" is polyphyletic as it includes birds and mammals.

Taxon 1 Species F

Taxon 3 is paraphyletic. It includes species A without including all of A's descendents. The traditional grouping of reptiles is paraphyletic because it does not include birds.

Taxon 2 Species Species G H

Species I

Species D

Species C

Taxon 3 Species Species J K

Species E

Species B

Taxon 1 is monophyletic as all the organisms are related to species B (the common ancestor). All the descendents of the first reptiles form a monophyletic group.

Species A

1. What does a phylogenetic tree show?

(a) Monophyletic:

(b) Polyphyletic:

(c) Paraphyletic: WEB

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3. Define the following:

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2. Why might a phylogenetic trees based on molecular differences be preferred over phylogenies based on appearance?

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163 Constructing Phylogenies Using Cladistics ancestor and all its descendents (i.e. it is monophyletic). Increasingly, cladistic methods rely on molecular data (e.g. DNA sequences) to determine phylogenies. Highly conserved DNA sequences are used because changes are likely to signal a significant evolutionary divergence. Cladograms may not always agree completely with phylogenies constructed using traditional methods but similarities in the trees indicate that the proposed relationships are likely to be correct.

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Key Idea: Cladograms are phylogenetic trees constructed on the basis of shared derived characteristics. A cladogram is a phylogenetic tree constructed using a taxonomic tool called cladistics. Cladistics groups organisms on the basis of their shared derived characters (features arising in an ancestor and shared by all its descendents) and ignores features that are not the result of shared ancestry. A clade, or branch on the tree, includes a common

Derived vs ancestral characters

When constructing cladograms, shared derived characters are used to separate the clades (branches on the tree). Using ancestral characters (those that arise in a species that is ancestral to more than one group) would result in distantly related organisms being grouped together and would not help to determine the evolutionary relationships within a clade. Whether or not a character is derived depends on the taxonomic level being considered. For example, a backbone is an ancestral character for mammals, but a derived character for vertebrates. Production of milk is a derived character shared by all mammals but no other taxa.

The backbone in a mammal, e.g. rat, is an ancestral character common to all vertebrate taxa. However, the production of milk from mammary glands is a derived character, shared by all mammals but no other taxa.

Constructing a simple cladogram

Jawless fish (outgroup)

Bony fish

Amphibians

Lizards

Birds

Mammals

Several different cladograms can be constructed from the same data. To determine the most likely relationships, the rule of parsimony is used. Parsimony assumes that the tree with the simplest explanation (the least number of evolutionary events) is most likely to show the correct evolutionary relationship.

Taxa

Comparative features

A table listing the features for comparison allows us to identify where we should make branches in the cladogram. An outgroup (one which is known to have no or little relationship to the other organisms) is used as a basis for comparison. The table (right) lists features shared by selected taxa. The outgroup (jawless fish) shares just one feature (vertebral column), so it gives a reference for comparison and the first branch of the cladogram. As the number of taxa in the table increases, the number of possible trees that could be drawn increases exponentially.

Vertebral column

Jaws

Four supporting limbs

Amniotic egg

Diapsid skull

Feathers

Hair

A possible cladogram for the data in the table is shown on the right. Its construction assumed that six evolutionary events took place (labelled as blue bars on the cladogram). If other cladograms were constructed, but involved more evolutionary events, the one shown would be assumed to be correct because it is the most parsimonious.

Jawless fish

Bony fish

Amphibians

Lizards

Birds

Mammals

Feathers

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Parsimony can lead to some confusion. Some evolutionary events have occurred multiple times. An example is the evolution of the four chambered heart, which occurred separately in both birds and mammals. The use of fossil evidence and DNA analysis can help to solve problems like this.

Hair

Diapsid skull

Amniotic egg

Limbs

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Jaws

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1. (a) Distinguish between a shared derived characteristic and a shared ancestral characteristic:

(b) Why are ancestral characteristics not useful in constructing evolutionary histories?

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2. What assumption is made when applying the rule of parsimony in constructing a cladogram?

3. Two possible phylogenetic trees constructed from the same character table are shown below. The numbers next to a blue bar represent an evolutionary event.

Species 7

Species 6

Species 5

Species 4

Species 3

Species 2

Species 1

Species 7

Phylogenetic tree 2

Species 6

Species 5

Species 4

(b) State your reason:

Species 3

Phylogenetic tree 1

Species 2

(a) Which tree is more likely to be correct?

Species 1

2

5

5

6

4

4

3

3 2

2

1

6

1

(c) Identify the event which has occurred twice in phylogenetic tree 2:

4. A phylogenetic tree is a hypothesis for an evolutionary history. How could you test it?

5. Use the shapes below to construct a cladogram that shows their phylogenetic relationships (hint: A is the outgroup).

C

D

E

F

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A

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164 Constructing a Cladogram

Key Idea: A table of selected characteristics can be organised systematically as a matrix and used to construct a cladogram.

1

2

3

4

5

6

7

8

9

10

11

12

13

Zebra-perch sea chub

0

0

0

0

0

0

0

0

0

0

0

0

0

Barred surfperch

1

0

0

0

0

0

0

0

0

1

1

0

0

Walleye surfperch

1

0

0

0

0

1

0

1

0

1

1

0

0

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Taxon

Black perch

1

1

1

0

0

0

0

0

0

0

0

1

0

Rainbow seaperch

1

1

1

0

0

0

0

0

0

0

0

1

0

Rubberlip surfperch

1

1

1

1

1

0

0

0

0

0

0

0

1

Pile surfperch

1

1

1

1

1

0

0

0

0

0

0

0

1

White seaperch

1

1

1

1

1

0

0

0

0

0

0

0

0

Shiner perch

1

1

1

1

1

1

0

0

0

0

0

0

0

Pink seaperch

1

1

1

1

1

1

1

1

0

0

0

0

0

Kelp perch

1

1

1

1

1

1

1

1

1

0

0

0

0

Reef perch

1

1

1

1

1

1

1

1

1

0

0

0

0

Selected characters for cladogram assembly 1. Viviparity (live bearing) 2. Males with flask organ 3. Orbit without bony front wall 4. Tail length 5. Body depth 6. Body size 7. Length of dorsal fin base 8. Eye diameter 9 Males with anal crescent 10 Pectoral bone with process 11. Length of dorsal sheath 12. Body mostly darkish 13. Flanks with large black bars

Juvenile surfperch

Steve Lonhart (SIMoN / MBNMS) PD NOAA

Character

Surfperches are viviparous (live bearing) and the females give birth to relatively well developed young. Some of the characters (below, left) relate to adaptations of the male for internal fertilisation. Others relate to deterring or detecting predators. In the matrix, characters are assigned a 0 or 1 depending on whether they represent the ancestral (0) or derived (1) state. This coding is common in cladistics because it allows the data to be analysed by computer.

Notes and working space

0 No 1 Yes 0 No 1 Yes 0 Yes 1 No 0 Short 1 Long 0 Deep 1 Narrow 0 Large 1 Small 0 Long 1 Short 0 Moderate 1 Large 0 No 1 Yes 0 No 1 Yes 0 Long 1 Short 0 No 1 Yes 0 No 1 Yes

1. This activity provides the taxa and character matrix for 11 genera of marine fishes in the family of surfperches. The outgroup given is a representative of a sister family of rudderfishes (zebra-perch sea chub), which are not live-bearing. Your task is to create the most parsimonious cladogram from the matrix of character states provided. To help you, we have organised the matrix with genera having the smallest blocks of derived character states (1) at the top following the outgroup representative. Use a separate sheet of graph paper, working from left to right to assemble your cladogram.

Identify the origin of derived character states with horizontal bars, as shown in the previous activity. CLUE: You should end up with 15 steps. Two derived character states arise twice independently. Staple your cladogram to this page.

2. (a) Why are the character states organised in a matrix?

(b) Why is it useful to designate the characters states as 0 (ancestral) or derived (1)?

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3. In the cladogram you have constructed for the surfperches, two characters have evolved twice independently: (a) Identify these two characters:

(b) What selection pressures do you think might have been important in the evolution of these two derived states?

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165 Master Genes and the Control of Development

Key Idea: Master genes control the expression of many other genes and regulate development. The same genes, or their homologues, are found in almost all eukaryotes and are highly conserved. Master genes regulate the activity of other genes in a coordinated way, shaping embryonic development by acting as a master switch, producing transcription factors that turn

Genetic switches

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Master genes control other genes

other genes on or off. Master genes have diverse roles; controlling the development of body parts, such as limbs and organs, controlling patterns and rates of growth, or orientation (e.g. front and back, top and bottom). Master genes are highly conserved gene sequences and they are found in almost all eukaryotes. Their highly conserved nature indicates their importance in the normal development of an organism.

The Distal-less gene is one of the important master body-building genes in the genetic tool kit. Switches in the Distal-less gene control expression in the embryo (E), larval legs (L), and wing (W) in flies and butterflies, but butterflies have also evolved an extra switch (S) to control eyespot development. Three switches in a fly

E

L

W

gene

A new switch, S, evolved in butterflies

E

L

W

S

gene

Once spots evolved, changes in Distal-less expression (through changes in the switch) produced more or fewer spots (below). An homologous gene called DLX is found in vertebrates. It is involved in formation and development of the skull, jaws, and face.

Master gene ‌

‌ switches other genes on or off in a specific sequence

Master genes, such as Hox, Pax, and BMP genes, regulate the expression of other genes by switching them on or off (above) via transcription factors. This is analogous to a director directing scenes in a play. If the actors are called to play a scene in the wrong order, the play will be presented differently.

A relatively small number of master genes are often called the genetic toolkit because the pattern in which they turn genes on or off produces the variation seen amongst organisms. The switching pattern varies not only between organisms but also in different regions of the body. For example, in a fly, the master sequence in the head region will switch on the genes for the development of antennae, but the genes to produce legs will be switched off.

Northern jungle queen butterfly

Buckeye butterfly

Silky owl butterfly

Master genes help explain evolutionary change

The study of developmental processes as a way to investigate ancestry is part of a powerful new field of evolutionary biology called evo-devo. Evo-devo provides some of the strongest evidence for the mechanisms of evolutionary change.

1. (a) What are master genes?

(b) What is their role?

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2. Using Distal-less as an example, explain how the activity of a master gene can produce differences in appearance?

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Key Idea: Variations in the expression of the master gene BMP4 produce the wide variety of phenotypes observed in the beaks of Darwin's finches and the jaws of cichlid fishes. Bone morphogenetic proteins (BMPs) regulate bone and cartilage growth in embryos. BMP4 is a signalling molecule (produced by a gene of the same name) that triggers a signal

transduction pathway resulting in the expression of the genes controlling craniofacial development. The diverse shapes seen in the beaks of Darwin's finches and in the jaws of African cichlid fishes are the result of different levels of BMP4 expression. They show how different morphologies can arise by changing gene expression, not the genes themselves.

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Depth

Beak shape is regulated by BMP4 and CaM1 expression

Depth

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166 The Role of Master Genes

The two extremes of beak shape in Darwin's finches are shown right. A wide range of beak shapes occur between these two extremes as a result of the different levels of expression of BMP4 and CaM1. The images below show some of the beak variation observed in Darwin's finches.

Wi

dth

Length

BMP4 (encoded by the BMP4 gene) controls beak width. Individuals expressing high levels of BMP4 early in development have deep, broad beaks, whereas individuals with low BMP4 expression developed narrower, shallower beaks.

The finches of the Galápagos Islands have adapted to occupy a wide range of different ecological niches, each specialising on different food sources. The shape of their beaks reflects this. Although several species of ground finch eat seeds, the size and hardness of the seeds they preferentially eat depends on their beak morphology. For example, the large, deep beak of the large ground finch allows it to break much harder, larger seeds than the smaller beaks of the two other ground finches.

Wi

dth

Length

CaM1 (encoded by the CaM1 gene) controls the length of the beak. When CaM1 is strongly expressed the beak tends to be more long and pointed than in individuals where CaM1 expression is lower.

The sharp-beaked finch (G. difficilis) feeds on a mixed diet of seeds and insects. Low BMP4 Low beak depth/width Low CaM Short beak

Compiled from various sources including: LF DeLeon et al, J. Evol. Biol, 27 (2014) "Darwin’s finches and their diet niches: the sympatric coexistence of imperfect generalists." and A Abzhanov et al. Nature 442 (2006) " The calmodulin pathway and evolution of elongated beak morphology in Darwin's finches."

Low/ moderate BMP4 Moderate beak depth/width

Moderate BMP4 Moderate beak depth/width

Early/high BMP4 High beak depth/width

High CaM Elongated beak

High CaM Elongated beak

Low CaM Short beak

Low CaM Short beak

Harvey Barrison cc 2.0

Low BMP4 Low beak depth/width

Finch photos all thanks to Prof. Jeff Podos unless otherwise stated

Two regulatory proteins, BMP4 and calmodulin (CaM1), are involved in controlling the shape of beaks in birds. Calmodulins are a group of calcium binding proteins, they have important roles in regulating a number of different protein targets, so are involved in regulating a variety of cell functions. In beak development, BMP4 controls width and CaM1 influences length.

Cactus seeds

Roger Culos Muséum de Toulouse CC3.0

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1 cm LINK

Large ground finch (G.magnirostris) Crushes large, hard seeds up to 2 cm.

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Cactus fruit

Large cactus finch (G. conirostris) Medium ground finch (G. fortis) Feeds on seeds, insects, and cacti. Crushes seeds up to 1.5 cm.

LINK

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Cactus finch (G. scandens) Probes cactus flowers and fruit.

165 147

LINK

67

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166

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Variations in cichlid jaw shape

Labeotropheus fuelleborni

Metriaclima zebra

The cichlids are a large, diverse family of perch-like fishes. They are particularly diverse in the African Great Lakes where adaptive radiation has resulted in ~1500 species exploiting a range of feeding niches. The morphology of the jaws and teeth in cichlids is associated with the level of BMP4 expressed during embryonic development. High levels of BMP4 are detected in the jaws of developing Labeotropheus fuelleborni. As a result, this species develops a short, robust lower jaw, with small closely spaced teeth, ideal for stripping algae from rocky surfaces. At the same stage of development, lower levels of BMP4 are expressed in Metriaclima zebra. The resulting jaw is more elongated, with large well spaced teeth. The M. zebra jaw is suited to sucking up loose material. Further evidence for the importance of BMP4 in jaw development has come from the manipulation of BMP4 expression in zebrafish (Danio rerio). Over-expression of BMP4 resulted in deep jaws forming.

1. (a) Describe the role of BMP4 in beak development:

(b) What role does CaM1 play in beak development?

2. The photos right show two species of GalĂĄpagos Island finches; the vegetarian finch (A) and the green warbler finch (B). Compare their beaks and suggest how the expression of BMP4 and CaM1 has likely differed between the two:

A

B

3. (a) What evidence is there that BMP4 plays a role in controlling the morphology of cichlid jaw development?

(b) How is jaw shape is related to diet in the two cichlid species above:

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Finch photos courtesy of Prof. Jeff Podos

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4. Explain how the regulation of gene expression has produced diverse morphologies in related species and explain the significance of this to their adaptive radiation:

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167 KEY TERMS AND IDEAS: Did You Get It?

1. Test your vocabulary by matching each term to its correct definition, as identified by its preceding letter code. BMP4

A A technique in molecular evolution that uses molecular change to deduce the

common ancestor

time in geologic history when two species or other taxa diverged. Can be used to establish phylogenies.

B The evolutionary history or genealogy of a group of organisms.

cladogram

C A diagram showing the evolutionary history of a group of organisms. D A technique used to determine the percentage similarity between the DNA of

master gene

E A single gene that controls the expression of many other genes in a

molecular clock

F DNA located in mitochondria.

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two organisms.

coordinated fashion.

G A master gene encoding a signalling protein involved in the regulation of bone

mtDNA

and cartilage development, especially in the craniofacial region.

H The individual from which all organisms in a taxon are directly descended.

phylogenetic tree

I A type of phylogenetic tree that groups organisms based on their shared derived characteristics.

phylogeny

2. Compare and contrast DNA hybridisation and DNA sequence comparison as methods for generating phylogenies:

3. The diagram left shows the evolutionary relationship of a group of birds based on DNA similarities:

DNA difference score

10

5

0

Flamingo

(a) Place an X to the last common ancestor of all the birds:

(b) How many years ago did storks diverge from vultures?

(c) What are the most closely related birds?

(d) What is the difference in DNA (score) between:

Ibis

Shoebill

Pelican

Stork

New World vulture

30

20

10

ii: Ibises and shoebills:

(e) Which of the birds is the least related to vultures?

0

Millions of years ago

4. African cichlid species (right) show great variation in the shape of their jaws. Describe how this variation arises:

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40

i: Storks and vultures:

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50

TEST


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Human change over time

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Unit 4 Outcome 1

Key terms

Primate characteristics

ape

Key knowledge

A. afarensis

Ardipithecus ramidus

Australopithecus spp.

c

1

Describe the shared characteristics that define primates, hominoids, and hominins. What features distinguish hominoids and hominins?

c

2

Appreciate that the classification of the primate order has had several revisions in recent decades to accommodate new genetic evidence.

bipedal (bipedalism) carrying angle

Major trends in hominin evolution

gracile

c

H. floresiensis H. habilis

168

Describe major trends in hominin evolution from the genus Australopithecus to the genus Homo. Include reference to structural (physical), functional, and cognitive changes and their significance to cultural evolution. Include:

172 - 184 192 193

i

The distinguishing characteristics of genus Australopithecus and the significance of A. afarensis (Lucy).

172 - 175

ii

The distinguishing characteristics of genus Homo and the biological and cultural evolution of species of Homo that illustrate the trend (H. habilis, H. erectus, H. neanderthalensis, H. sapiens).

172 - 175

3

Homo

H. erectus

169 - 171

Activity number

Key knowledge

hominin

hominoid

Activity number

H. neanderthalensis H. sapiens Mesolithic

c

4

Relate the evolution of hominins to changes in habitat and a shift in resources.

172 176

Neolithic

c

5

Describe the selection pressures on early hominins and the benefits of reducing body hair and adopting bipedalism as a form of locomotion.

176 - 177

primate

Palaeolithic prehensile

prognathic robust

valgus angle

The complexity of the human fossil record Key knowledge

Activity number 184

i

the significance of the Denisova hominin to our understanding of human lineages and the genetics of modern populations

186

c

ii

the evidence for interbreeding between Homo sapiens and neanderthals

187

c

iii

the importance of Ardipithecus to our understanding of the events in early hominin evolution

185

c

iv

the implications of the fossils on the island of Flores (H. floresiensis)

188

c

v

the significance of finds that are not datable (the Rising Star hominin, H. naledi).

189

c

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6

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Recognise the human fossil record as an example to illustrate why and how classifications can be questioned, refined, and reevaluated, to include at least i and ii of the following examples (iii-v optional):

c


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168 Primate Classification

Key Idea: Humans are primates, with genetic evidence showing chimpanzees to be our closest living relatives. The classification of primates has been considerably revised several times in the last few decades. Classification should reflect evolutionary history, and the modern classifications of primates are supported by genetic, as well as morphological, evidence of phylogeny. Current classifications favour two primate suborders: the Strepsirrhini (lemurs and lorises)

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and the Haplorhini (tarsiers, monkeys, apes, and humans). Within the haplorhines, the classification of the apes has been particularly controversial, and heavily influenced by the historical perception that humans were special and worthy of their own family. Based on genetic evidence, chimps and gorillas are more closely related to humans than to orangutans, and chimps are more closely related to humans than they are to gorillas.

Superfamily Hominoidea

Family Hylobatidae:

Family Hominidae:

Subfamily Ponginae

Phylogenetic tree of the anthropoids

(various gibbon species)

Genus Pongo:

Pongo pygmaeus Bornean orangutan Pongo abelii

Sumatran orangutan

The diagram below presents a classification of the anthropoids (the hominoids and monkeys) based on genetic differences. The percentages next to each of the points where a split occurs indicates the amount of difference in the total genetic makeup (genomes) of the two groups being considered, e.g. .the genome of the gibbons compared to the rest of the apes (orangutans, gorillas, chimpanzees) and humans differs by 5.7%. A large genetic difference between any two groups implies that they are distantly related, whereas small genetic differences suggest they share a recent common ancestor.

Subfamily Homininae

Tribe Gorillini

Genus Gorilla:

Gorilla gorilla

Western gorilla

Gorilla beringei

Eastern gorilla

Human

1.4%

Tribe Hominini

Bonobo

Subtribe Panina Genus Pan:

Pan troglodytes

Common chimpanzee

1.8%

Common chimpanzee

Pan paniscus Bonobo

Western gorilla

Subtribe Hominina

Genus Homo:

Eastern lowland gorilla

3.6%

Homo sapiens Human

Eastern mountain gorilla Bornean orangutan

5.7%

Sumatran orangutan

7.9%

Gibbon

13.0%

Old World monkey

New World monkey

40

30

20

10

0

Millions of years ago

1. According to the diagram above, which shows relatedness according to genetic similarity: (a) Which hominoid group is most closely related to the two chimpanzee species?

(b) Name the two chimpanzee species:

2. From the diagram, determine how long ago:

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(a) The two species of chimpanzee split from a common ancestor:

(b) The chimpanzees split from the line to humans:

(c) The African apes (and humans) split from the Asian apes (orangutans and gibbons):

3. Circle on the diagram the point where the African (Old World) monkeys split from the American (New World) monkeys: LINK

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LINK

WEB

169 162 168

KNOW


169 General Primate Characteristics

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242

Key Idea: Primates exhibit unique, but quite generalised, morphological, physiological, and behavioural features. Morphological features include five digits on the hands and

feet, physiological features include a longer gestation than other mammals, and behavioural features include prolonged infant dependency and keen social behaviour.

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The gestation (pregnancy period) in primates is longer than most other mammals. Primates typically have one young per pregnancy (below). Infancy is prolonged with longer periods of infant dependency and a large parental investment in each offspring. This nurturing increases the survival rate of the young and allows cultural development.

Primates generally live longer relative to most other mammals and there is a greater dependency on highly flexible learned behaviour. Primates tend to be highly sociable (above). Unusually for mammals, adult males of many primate species often associate permanently with the group. The brain is large and generally more complex than in other mammals. Vision is important, the visual areas of the brain are enhanced. Well developed binocular, stereoscopic vision provides overlapping visual fields and good depth perception. Colour vision is probably present in all primates, except specialised nocturnal forms.

RA

Primates have a tendency toward erectness, particularly in the upper body, as seen in the gorilla, above. This tendency is associated with sitting, standing, leaping, and (in some) walking. WEB

KNOW

LINK

Gibbons are the smallest of the apes and are specialised to use brachiation (a technique of under-branch swinging), in combination with rapid climbing, midair leaps, and bipedal running, to move quickly through the forest.

LINK

169 170 171

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Primates have a generalised dental pattern particularly in the back teeth. Unspecialised teeth enabled primates to adopt a flexible omnivorous diet.

Monkeys walk quadrupedally on the palms of their hands and the soles of their feet. In the trees, they walk along the tops of branches, gripping them with their hands and feet.

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Chimpanzees and gorillas spend more time out of trees than do either of the Asian apes. The chimpanzee above shows typical knuckle-walking behaviour. The relatively long arms facilitate this mode of locomotion.

A trend towards a reduced snout and flattened face and reduced olfactory regions in the brain. Baboons go against this trend, with a secondary increase in muzzle length.

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1. On the diagram of the capuchin below, briefly describe the general physical characteristics of all primates as indicated: The primate pictured is a white-fronted capuchin monkey (Cebus albifrons) from northern South America. These monkeys inhabit the mid-canopy deciduous, gallery forests.

Brain size and specialisation:

Vision:

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Collarbone:

Face shape and snout:

Teeth shape and dental arrangement:

Posture:

Hands and feet:

Limb joints:

Reproduction:

Social organisation:

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2. Humans belong to the order Primates. Describe the features of humans that characterise their primate heritage:


170 The Primate Hand

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244

Key Idea: Primates have grasping hands, but the human hand is particularly developed with respect to dexterity and the manipulation of objects. Primates have a grasping hand. They are able to pick things

up, hold, and manipulate them, although the degree to which a primate can do this depends on the species. Humans have a highly advanced ability to manipulate objects with their hands because the thumb is very long relative to the hand.

The fingers have end tactile pads that contain huge numbers of nerve endings, producing a highly sensitive surface.

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Index finger and little finger are able to pivot and move towards each other, allowing the fingers to form around small objects.

Large muscles, especially around the thumb, produce a powerful grip.

Nails are found on at least some digits in all modern primates.

Flexible joints in the hand allow it to flex, increasing dexterity.

Highly mobile thumb, able to touch all other fingers on the hand.

Chimpanzees are very capable tool users, but their dexterity is limited by the length of their hand compared to the thumb and the rigid wrist bones that limit wrist rotation.

LBS

New research suggests manipulating objects may not have been the only important factor in the evolution of the human hand. The human hand shape is one of the only configurations possible that maintains dexterity while allowing the hand to form a fist. Although a punch with a closed fist produces the same force as a slap with an open palm, a punch delivers the force to a smaller area, producing a much greater impact and potential for damage to an opponent.

Grips of the human hand

Power grip

Power grip

Precision grip

All photos LBS

Precision/ power grip

1. Name two possible selective pressures acting on the human hand:

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2. Explain why being able to manipulate objects is an evolutionary advantage:

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LINK

LINK

169 176 178

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3. Identify two features that make the human hand so dexterous and compare them to a chimpanzee hand:

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171 Hominoids and Hominins

Key Idea: The hominins include modern humans and their extinct ancestors. Hominins are part of a larger superfamily, the hominoids, which also includes the apes. The hominoids (apes as well as humans and their ancestors) are large, tailless primates, with bony eye ridges and flattened

noses. Hominins (humans and their extinct closest ancestors) are a subtribe within this larger taxon identified by human features including dentition and brain size. The older taxon hominid is now a collective term encompassing the great apes and hominins and no longer refers just to humans.

Family Hominidae (hominids)

Superfamily Hominoidea

Characteristic features of the hominids: • Large and sexually dimorphic • Most predominately quadrupedal • Most omnivorous • Typical ape-like dentition but teeth large in gorillas and small in humans • Complex social behaviour

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Hominoid features (lesser apes & hominids) • • • •

No tail Semi-erect or fully erect posture Broad chest, pelvis, and shoulders Relatively long arms and mobile shoulder joints • Larger brain

Subfamily Ponginae

Family Hylobatidae

Subfamily Homininae

Features possessed to varying degrees by subfamily Homininae: • Partially or habitually bipedal and ground dwelling • Large cerebral cortex • Reduced canines • Highly sensitive skin, body hair reduced

Orangutan

Siamang

Tribe Gorillini: Gorillas

Gibbon

Characteristic features of the lesser apes: • • • •

Tribe Hominini: Humans, their ancestors, and chimpanzees

Subtribe Hominina Humans & their ancestors

All found in Southeast Asia Long forearms with hook-like fingers specialised for brachiation Pads on the rump (ischial callosities) Arboreal; sleep on tree branches and do not build nests

Subtribe Australopithecina Australopithecines (extinct)

Subtribe Panina Chimpanzees

Hominins

1. Use the information above to complete the phylogeny of Hominoidea:

Hominoidea

Superfamily

Family

Subfamily

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Tribe

2. What are the differences between hominoids and hominins?

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Subtribe

LINK

168

KNOW


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172 Trends in Hominin Evolution: Overview

Key Idea: The hominin fossil record shows clear evolutionary trends towards bipedalism, increased brain size, increased height, and increased technical ability. The diagram below and opposite shows a consensus view of the trends in hominin evolution over time. Only the five species representative of the general trends are shown here.

Australopithecus afarensis

The early australopithecines were almost certainly ancestral to Homo habilis, which was ancestral to modern humans. Some populations of Homo erectus migrated out of Africa, eventually giving rise to populations of Homo in the Middle East and Europe. Neanderthals eventually evolved in Western Europe and modern humans in Africa.

Homo erectus

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Homo habilis

Brain capacity 1016 cm3

Brain capacity 552 cm3

Brain capacity 457 cm3

3.9 million 2.9 million years ago

2.8 million 1.5 million years ago

1.9 million 600,000 years ago

179 cm

130 cm

110 cm

0

CC

4.

u

co

es

s en

rD

e di

Di

Oldowan tools

Australopithecus afarensis

Homo habilis

Acheulean tools

Homo erectus

(a) Angle of the face:

(b) Size of the brain and skull:

(c) Height and stance:

(d) Skill at tool making:

WEB

KNOW

LINK

LINK

LINK

172 173 174 175

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1. Describe the general trends in the following features:

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247 Climate and environmental changes

Forested

Fluctuating glacial/ interglacials

Open grassland

Warm

Cool

4 mya

2 mya

Present

Homo neanderthalensis

PR E O V N IE LY W

Homo sapiens

Brain capacity 1512 cm3

Brain capacity 1335 cm3

500,000 - 40,000 years ago

200,000 years ago - present

183 cm

Didier Descouens CC 4.0

170 cm

Mousterian tools

Neolithic tools

Homo sapiens

Homo heidelbergensis (not shown)

Homo neanderthalensis and Homo sapiens lived at roughly the same time. However H. sapiens evolved in warmer Africa and H. neanderthalensis evolved in cooler Europe. They may have overlapped in space for a few thousand years as H. sapiens migrated out of Africa.

Homo neanderthalensis became extinct about 30,000 years ago.

Homo neanderthalensis

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2. What was happening to the climate and environment as human ancestors evolved?

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3. The skulls of Homo neanderthalensis and Homo sapiens are similarly sized. Describe the main differences between them:


248

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173 Trends in Skull Anatomy

Key Idea: The trend in hominin skulls has been towards a greater volume, flatter face, and more gracile features. The shape of the modern human skull is quite different from its ancestors and that of Neanderthals. The human skull has a very high forehead and domed skull, whereas early ancestors had flatter foreheads and more elongated skulls.

Modern humans have rather gracile skulls compared to their ancestors. The cheek bones and jaw are both smaller and the brow ridges are much reduced relative to earlier hominins. These changes tend to reflect a change in diet to one that requires less chewing (e.g. from tough vegetable matter to a greater amount of (cooked) meat).

Skull features

3. Braincase (a) Shape of forehead (slope, height). (b) Rear view: where is skull the widest, low down or high up? Shape: pentagonal, rounded, bell-shaped? (c) Presence of crests: Nuchal crest for neck muscles, Sagittal crest for jaw muscles. (d) Shape of occipital region (back of skull) when viewed from the side: presence of bun? (e) Dorsal (top) view: where is the skull widest (rear, middle ear, etc.)? (f) Position of foramen magnum (opening at base of skull connected to spine).

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The skulls show in this activity are representative of the many hominins both modern and prehistoric. The features mentioned below and shown in the diagram below are features that should be noted when looking at trends in skull evolution. 1. Face (a) Size of the face compared to the braincase. (b) Degree of prognathism (snout or muzzle development) of the jaw and mid face (mid-face projection). (c) Development of brow ridges (supraorbital tori): size, thickness, arching. (d) Size of cheek region. 2. Jaws (mandible) (a) Size and thickness of lower jaw. (b) Degree of curvature of dental arcade (tooth row). (c) Presence or absence of chin.

Sagittal crest present? (site of attachment for jaw muscles)

Shape and slope of forehead

Brow ridge development?

Brain case: size and shape

Facial angle

Nuchal crest present? (site of attachment for neck muscles)

Size and shape of zygomatic arch (cheek bones)

Size of biting front teeth (incisors), canines and molars

Position of the foramen magnum (hole at the base of the skull that joins on to the spine)

Degree of prognathism (snout or muzzle)

Size of mandible (jaw bone)

Diastema (gap) between incisors and canines present of absent?

Chin present?

1. For each of the hominin species describe the features of the skull. Use the diagram above as a guide.

(b) Homo habilis:

(c) Homo erectus:

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LINK

173 172

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(a) Australopithecus afarensis:

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249

Australopithecus afarensis

Homo habilis

Homo erectus Heavy brow ridge

PR E O V N IE LY W

Sloping forehead

Large teeth, including large canines

Angle/position of foramen magnum (FM)

Occipital bun

FM

FM

No chin

Prominent cheek bones

Homo neanderthalensis

and are generally thought of as being adapted for the cooler conditions found there. However examination of the nasal cavity finds it does not fit the general rule in mammals that in cold climates there is usually a reduction in the size of the nose and nasal cavities. The Neanderthal nose size is therefore a bit of an oddity and may be linked simply to the degree of facial projection. Also the internal sinuses are small and do follow the rule for cold climates.

ffThe skull features a large occipital 'bun' which may

reflect an enlarged occipital lobe. The occipital lobe is involved with visual processing. It could also reflect a larger cerebellum, which is involved in the coordination of movement and spatial information.

(d) Homo neanderthalensis:

(e) Homo sapiens:

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FM

ffThe modern human skull sports a high vertical forehead

and large domed skull. This reflects an increase in the size of the frontal lobe of the cerebral cortex.

ffThere is no brow ridge and the facial and cheek bones have

all been reduced in size. The jaw is smaller relative to the skull than in any other hominin, as are the cheek bones and teeth. This may reflect a shift in diet to food that required less chewing (less powerful musculature would be needed so muscle attachments can be less robust).

ffThere is also no nuchal ridge as the skull is now balanced

directly above the spine so only small muscles are required to hold it upright. There is a prominent chin which acts as a buttress (support) for the small jaw.

N AS OT SR F OO OR M US E

ffNeanderthals evolved in Southern and Western Europe

Homo sapiens

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FM

Large jaw


174 Trends in Brain Volume

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250

such as elephants and whales, with brain volumes greater than ourselves and yet they are not considered to be as intelligent. It appears that what is more important is the relative brain size (brain size relative to body size). Modern humans have a brain volume three times larger than that predicted for an ape with our body size. The organisation of the brain is also important. Apart from the highly developed cerebrum, two areas of the brain associated with communication have also become highly developed in modern humans: Broca’s area, concerned with speech, and Wernicke’s area, concerned with comprehension of language.

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Key Idea: The evolution of a large brain was crucial to our development of language, technology, and culture. The human brain is responsible for our unique human behavioural qualities. It makes up just 2% of our body weight, but demands about 20% of the body’s metabolic energy at rest. This makes the brain an expensive organ to maintain. The selection pressures for increased brain size must have been considerable for additional energy to be made available. The normal human adult brain averages around 1330 cc, but ranges in size between 1000 and 2000 cc. But intelligence is not just a function of brain size. There are large mammals, Growth in brain size in humans and chimpanzees

Brain volume for hominin species

Hominin species

1500

Average brain vo lume (cm3)

Australopithecus afarensis

3.5

440

Australopithecus africanus

2.5

450

Paranthropus robustus

2.0

520

Paranthropus boisei

1.5

515

Homo rudolfensis

2.0

700

700

Homo habilis

575 560

500

Homo naledi Homo ergaster

1.8 ?

1300

Brain size (cm3)

Years ago (mya)

Humans

1100

900

Chimpanzees

300

100

0

2

4

6

8

10

12

14

16

18

20

Age (years)

Figure above: In most primates, including chimpanzees, brain growth, relative to body size, slows markedly after birth while body growth continues. In human infants, the slowing of brain growth does not occur until more than a year after birth, which results in larger brain masses for humans than for chimpanzees at any given age (or body weight).

1.8

800

Homo erectus

0.5

1100

Homo heidelbergensis

0.2

1250

Homo neanderthalensis

0.05

1550

Homo floresiensis

0.05

380

Homo sapiens

0.08

1350

Table above: A generalised summary of the changes in estimated brain volume recorded from the fossil remains of hominins. The dates for each species are generally the middle of their time range for long-lived species or at the beginning of their time range for short-lived species.

1. Plot the data in the table on the estimated Brain Volume for Hominin Species (above) onto the graph below. Changes in hominin brain volume over time

1600

Mean volume of 1335 cm3 for living humans

3

Estimated brain volume (cm )

1400

1200

1000

800

600

Mean volume of 400 cm3 for chimpanzees

200

0 4.0

3.0

2.0

Millions of years ago

1.0

N AS OT SR F OO OR M US E

400

0

LINK

KNOW

172

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2. There were two ‘bursts’ (sudden increases) of brain expansion during human evolution. Indicate on the graph you have plotted where you think these two events occurred.

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Brain size vs body height in hominins

Frontal lobe

2000

Homo sapiens

1000

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Brain volume (cm3)

1500

Erectines

500

Broca’s area

Controls the muscles of the lips, jaw, tongue, soft palate, and vocal cords during speech.

Homo naledi

Australopithecines

Homo floresiensis

0

1.00

1.25

1.50

1.75

Cerebellum

Wernicke’s area

The area concerned with the understanding of spoken words.

2.00

Height (m)

Brain size can be correlated with body height in hominins. Three distinct clusters emerge, indicating three phases of evolutionary development. Homo floresiensis, found on the Indonesian island of Flores, clearly falls outside these clusters. Its brain size to body size ratio is similar to that of the Australopithecines, but key aspects of its morphology, such as its small canine teeth and organisation of the brain, identify it as Homo. In addition, the Flores finds were associated with relatively advanced stone tools.

Modern human brain

The human brain is very large for a primate of our size, but this may not be as important as its internal organisation. The most important specialisation of the human brain is the capacity for language: a result of the development of Wernicke’s and Broca’s areas. Specific differences associated with the left and right hemispheres of the brain are associated with these specialisations.

3. Why is brain volume alone not a reliable indicator of intelligence?

4. Explain the significance of the high energy requirement of a relatively large brain:

5. Comment on the significance of the brain/body size growth curve in humans compared with other primates:

6. (a) With respect to the brain size: body size ratio, comment on the position of H. floresiensis with respect to other hominins:

(b) Comment on the significance of the Flores finds:

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7. There is no firm date for H. naledi. Based on your plot opposite, what approximate date would your assign to this fossil?

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8. Describe a likely selection pressure for the evolution of increased brain size in early humans:


175 Trends in Dentition

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Key Idea: Changes in dentition (the type, number, and arrangement of teeth) and jaw structure of our hominin ancestors can reveal information about our evolution. During early hominin evolution teeth (especially the molars) and jaws tended to be large. The paranthropines are the extreme example of this trend. Their diet of coarse vegetation required very large and powerful jaws and molars. During the

course of hominin evolution, there was a general trend for a reduction in the size of the teeth tooth and jaw. This was a likely consequence of including a greater proportion of cooked foods, which required less chewing, in the diet. The teeth of modern humans are relatively small and generalised, reflecting an omnivorous diet of mainly processed (e.g. cooked) foods.

Late hominins

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Early hominins

Australopithecus afarensis • • • • •

Relatively large canine teeth Relatively large jaw V-shaped dental arcade Thin tooth enamel Diet probably consisted of fruits with some tougher material

Homo habilis

• Reduced canine teeth • Large molars and incisors • Dental arcade more like H. sapiens but still intermediate • Thick tooth enamel • Diet probably included vegetable matter and some meat (probably from scavenging)

Homo erectus

• • • • • •

Thick jaw bones No chin Relatively large molars Parabolic dental arcade Thick tooth enamel Diet probably included vegetable material and a large proportion of meat

Homo sapiens

• Shortened jaw, allows large bite force to be generated with little effort • Chin reinforces jaw, but leaves room for tongue muscles • Thick tooth enamel • Small molars adapted to chewing cooked and soft food • Parabolic dental arcade

Dental formulae all follow: I-2, C-1, P-2, M-3

Adaptations to a coarse diet

Paranthropus boisei had jaws and teeth adapted to a diet of very coarse vegetation and hard seeds. Their jaws produced a massive bite force of 2161 newtons, which helped to break food up. A modern human's maximum bite force is 777 newtons.

Teeth had a very thick coating of enamel to protect them.

Massive molars and premolars aided effective grinding action.

In many primates, the canine teeth are used in behavioural and social interactions, especially in species which show marked sexual dimorphism. Threat gestures, such as yawning (above), help maintain social order.

Reduced size of canines permitted rotatory action, helping to grind coarse food up. The reduced size of the incisors provides more room for molars.

The L shape of the jaw and the position of its joint allows the molars and premolars to meet at the same time giving an effective chewing action.

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2. What is one possible purpose of the chin in modern humans?

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1. Describe the general trend in the evolution of hominin teeth:

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176 Bipedalism and Nakedness both evolutionary responses to the changing climate of East Africa about 7-3 mya. However, a 2009 analysis of the 4.4 mya Ardipithecus fossils indicates that these very early hominins were still primarily forest dwellers, so any current hypotheses must account for the emergence of bipedalism in a forested environment. The Ardipithecus finds indicate that bipedalism was strongly associated with provisioning, and was later reinforced by a move into less forested habitats as savannah became established throughout Africa in the later Miocene.

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Key Idea: Bipedalism provided advantages such as better provisioning, greater safety, and greater efficiencies in locomotion and thermoregulation. The first major step in the evolution of humans as a distinct group from apes was their ability to adopt the habitually upright stance we call bipedalism. Closely linked to this shift was the reduction in body hair. A number of selection pressures for hair reduction are described below (left). Early studies suggested that bipedalism and hair reduction were

Hair reduction

Retention of head hair Hair on the head and shoulders has been retained to reflect and radiate heat before it reaches exposed skin.

Parasite control A reduction in body hair would have made it easier to control external parasites such as fleas and lice. This would have been increasingly important when early hominins began to use a 'home base'. Many external parasites need to complete their life cycle at a single location so that hatching eggs can reinfect their host.

Thermoregulation About 3 mya, the vegetation patterns in East Africa began to favour open grasslands, with fewer forested areas. This environment would have provided fewer opportunities for shelter from the sun, creating a selection pressure for the refinement of several thermoregulatory mechanisms.

Shorter, finer hairs (not hair loss) in early hominins would have allowed greater heat loss via radiation from the skin surface. Well developed sweat glands in humans enable heat loss at 700 watts m-2 of skin (greater than any other mammal).

Bipedalism

Seeing over the grass An upright posture may have helped early hominins to see predators or locate carcasses at a distance. Carrying offspring Walking upright enabled early hominins to carry their offspring, so the family group could move together. Provisioning as a selection pressure The ability to carry food while walking seems to have been important in the initial development of bipedalism. Females would have favoured males able to provide energy-rich foods, which would improve offspring survival and increase reproductive rate. The ability to carry food from its source to a place of safety would have had a great survival advantage. Efficient locomotion Once bipedalism was established, changing habitats would have provided selection pressure for greater efficiency. Being able to move across the growing savannah without expending large amounts of energy would have offered a great survival advantage. Holding tools and weapons Tool use was probably a consequence of bipedalism, rather than a cause. Upright walking appears to have been established well before the development of hunting in early hominids. Thermoregulation Upright walking exposes 60% less surface area to the sun at midday and there is greater air flow across the body when it is lifted higher off the ground.

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1. What advantages might an early human ancestor have gained by adopting a bipedal stance?

2. (a) What selection pressures are likely to have been important in the evolution of bipedalism initially?

(b) What environmental changes could have reinforced the advantages of bipedalism to human ancestors?

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177 Adaptations for Bipedalism

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Key Idea: Important changes in the skeleton are associated with the move to bipedal locomotion in early hominins. The reconstruction of Lucy (Australopithecus afarensis), opposite, shows the skeletal features of an early bipedal hominin. Lucy still possessed ape-like features but she was a fully-bipedal hominin with all the adaptations associated

with bipedal locomotion. Although there is no doubt that Lucy was habitually bipedal, a number of skeletal features suggest that tree climbing was still an important part of this hominin's niche, perhaps associated with escape, security, or foraging. A. afarensis is an important link between the quadrupedal locomotion of apes and bipedalism in hominins.

Chimpanzee

Human

Broad, basin-like pelvis

Long narrow pelvis

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Gluteus muscles prevent tilting when the opposite leg is off the ground.

Femoral head angled and strengthened

Foramen magnum (FM) toward the back of skull

The carrying (valgus) angle ensures the knee is brought under the body during walking.

Foramen magnum (FM) further forward so the skull balances on the spine S-shaped spine acts like a spring

Quadruped: Straight spine and rear-ward FM, femur is at right angles to knee so an upright stance is less stable.

Biped: S-shaped spine and forward FM. Femur (thigh) is angled out from knee (the carrying angle). A longer femur provides a longer, more efficient stride.

Chimpanzee

Human

End of femur at the knee joint

End of femur at the knee joint

Australopithecine

End of femur at the knee joint

Bony buttress

Lateral condyle

Inner (medial) condyle

Chimpanzee foot

Human foot

Lighter shading represents points of contact with the ground

Lighter shading represents points of contact with the ground

Australopithecine footprints

Direction of weight transmission during walking

Curved toe bones

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The human foot is adapted as a weight bearing platform rather than a grasping structure. The toes are reduced relative to those of chimpanzees. The foot is arched so transmits weight from the heel, along the outside of the foot, across the ball and through the big toe. This weight transference conserves energy during locomotion.

Heel bone missing from fossil

Foot bones (OH8) from Bed I at Olduvai Gorge

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The foot of a chimpanzee has relatively long, curved toes, with an opposable big toe adapted from grasping but ill-suited to upright walking. The foot transmits weight from the heel, along the outside of the foot, and then through the middle toes.

Big toe aligned with other toes (not opposable)

The australopithecine foot had an aligned big toe, as in humans, making it difficult if not impossible to grasp branches with the hindlimbs. The heel bones that have been found also indicate habitual bipedalism. Computer simulations suggest that A. afarensis could walk like humans but could not have walked like a chimpanzee.

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Big toe diverges (well separated from other toes

Large heel bears increased weight

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255 Lucy’s bipedal features

Shape of the tooth row (dental arcade) is half way between the straight-sided U-shape of an ape jaw and the more rounded, parabolic shape of a human jaw.

The foramen magnum was much further forward than in apes and much closer to the position in humans.

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Lucy’s* ape-like characteristics

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Shoulder joint orientated towards the head, similar to the orientation in chimpanzees and other arboreal quadrupeds.

Lucy’s lumbar vertebrae were broad for effective weight transmission from the upper body to the pelvis. The australopithecine spine had an S shaped curvature, similar to that of modern humans.

Funnel-shaped chest (thorax).

Lucy’s pelvis was broad and basin shaped, similar to a human pelvis. It would have supported the upper body when upright.

Lucy’s limbs showed human-like features consistent with bipedalism. The femoral (valgus) angle was similar to humans, bringing the knees under the body.

Arms relatively long compared to legs. Highly mobile wrist.

Curved finger bones.

Butressing of the knee was more similar to humans than to apes.

Relatively short legs.

Highly mobile ankle joint.

Toes are long and curved.

Arched feet, wide heels, and big toes aligned with the other toes and not opposable.

Redrawn from a photograph by © David L. Brill 1985

*Lucy is the name given to a specimen of Australopithecus afarensis

1. Describe features of each of the following in A. afarensis and explain what they tell us about how this animal moved:

(a) Foot:

(b) Pelvis:

(c) Lower limb:

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2. What does the retention of ape-like characteristics tell us about the niche of A. afarensis?

3. Compare the position of the foramen magnum in a chimpanzee and A. afarensis and comment on its significance:

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4. Imagine you are on an expedition to a well known hominin fossil site in east Africa. Describe a part of a hominin fossil skeleton that you would wish to find that would be ideal in clearly indicating bipedalism:


178 Cultural Evolution

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Key Idea: Cultural evolution is a term used to describe the transmission of and changes to knowledge and ideas over successive generations. In addition to the physical evolution of humans, ideas and

behaviours also evolved as they were learned and passed on to offspring. This non-genetic means of adaptation, called cultural evolution, further enhanced the success of early humans.

Environmental forces

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Over many millions of years, the evolution of human ancestors has been directed by natural selection. Environmental forces such as climate change, food supply, and predators, acted on the gene pool.

Climate change

The climate became drier and the forests which were the homes of the earlier primates gradually disappeared. This reduced shelter and meant that traditional food sources became scarce or disappeared. New food resources had to be investigated.

Resulting physical features

In response to these selective pressures humans evolved an upright stance with the head balanced on the spine and a large brain capable of learning, planning and passing on ideas. An upright stance freed the hands to grasp and manipulate objects in a very sensitive and precise way.

Fierce predators

Predators made a ground dwelling lifestyle dangerous. Early humans would have to develop solutions to protect themselves from attack.

Adopted niche

Predominantly ground living, opportunist/ scavenger. Able to exploit a number of varied habitats and utilise a range of food resources.

Cultural forces

The unique combination of brain and specialised physical features allowed early humans to learn from others and manipulate their environment to begin changing it to suit themselves.

Natural history intelligence

Social intelligence

Technical intelligence

Being able to predict, using current observations, the habits of potential game, the rhythms of the seasons, and the geography of the landscape (e.g. location of water sources and caves).

Language to communicate ideas, plan survival strategies, and coordinate group activities such as resource gathering and hunting of increasingly larger game. Group bonding behaviour improves survival opportunities for members.

Producing artefacts from mental templates required an understanding of abstract ideas and physical processes: the fracturing behaviour of stone, angles of striking stone and how hard to strike, and the trajectory of a thrown projectile.

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1. Explain what is meant by cultural evolution:

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The modern human mind

Creating artefacts and images with symbolic meaning as a means of communication. Using knowledge of animal habits, tools, advanced planning and communication to coordinate the hunting of large game.

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179 Trends in Palaeolithic Tool Cultures

Key Idea: The development of stone tools is a defining characteristic of Homo. Particular, identifiable types can be matched with specific time periods and species. The Palaeolithic (Old Stone Age) is a period of early cultural development spanning the emergence of the first stone tools

about 3.3 mya in eastern Africa, until the development of sophisticated tool kits in the Mesolithic (Middle Stone Age) about 10,000 ya. These tool cultures are known mostly by their stone implements. While other materials, such as wood, were probably also used, they did not preserve well.

Timeline of stone tool technologies through the Paleolithic

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Oldowan

3 mya

Oldowan (pebble) tool culture

Probably made by Homo habilis, these were crudely fashioned river-worn pebbles. A minimum number of flakes were knocked off from several angles to produce a core with a cutting edge (e.g. chopper, discoid, polyhedron). Although the cores may have been used as tools, it is known that the sharp flakes were also useful in cutting.

Mousterian

Upper Paleolithic

2 mya

1 mya

Flakes removed from one side only

Flakes removed from two sides

Cores

Chopper

Proto-biface

Acheulean tool culture

Upper Palaeolithic tool culture

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Polyhedron

Discoid

Biface shape: bulges outwards on both sides and has tear-drop shape

The core is the tool

Side-on view

Hand axe

Mousterian tool culture

There was a rather sudden increase in the sophistication of tool making about 35,000 to 40,000 years ago. Both the modern Homo sapiens and the last of the Neanderthals produced flint tools of much finer workmanship using a technique called punch blade. Long, thin flakes are removed and shaped into different tool types. A number of European sub-cultures, e.g. Solutrean, emerged. Other material such as bone, ivory and antler became increasingly utilised to produce very fine tools such as needles.

Flakes removed from all sides

Flakes (not shown) and the cores are used as tools

Made by H. erectus and archaic H. sapiens, these tools were typically 'tear drop' in shape and were crafted with a slight bulge on each broad surface (a bi-face). They ranged in size and are often referred to as hand axes although it is not clearly understood how they were used. They differ from the pebble tools in that there appears to be a standard design and each tool is manufactured using a great many more blows to remove flakes.

Made by Neanderthals and more refined than Acheulean tools. Flint became commonly used. This stone would chip in a predictable way when struck with another hard object so finer workmanship was possible. A particular technique from this period is known as the Levallois method. It involves the preparation of a core and striking off a large oval flake which is then retouched on one surface only (see the photograph on the right; the retouched surface is visible).

10,000 ya

Cleaver

Levallois scraper

Levallois method

Flake

Core

Handaxe from Le Moustier France. Flint, 8.5 cm.

Side scraper

Punch blade method

Solutrean blade

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3.3 mya

Acheulean

Bone needle

Core

Finely worked edge

Burin

Throwing stick

Flake

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Oldest dated stone tools

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D B

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A

C

1. Name the culture associated with each of the tools above (A-D) and describe the features that help identify them:

(a) Tool A culture:

(b) Tool B culture:

(c) Tool C culture:

(d) Tool D culture:

2. Identify the hominin species associated with, and the approximate time period for, each of the tool cultures below:

(a) Oldowan:

(b) Acheulian:

(c) Mousterian:

(d) Upper Palaeolithic:

3. Describe the general trends in the design of the stone tool from Oldowan to Upper Palaeolithic cultures:

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4. The tools that are recovered from early human prehistoric sites are almost invariably stone, bone or ivory. Explain why tools made from other materials are almost never recovered from these sites:

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5. Name the materials used to make tools in the Upper Palaeolithic culture that were seldom used in earlier cultures:

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180 Fire

Key Idea: The controlled used of fire by early humans allowed them to eat a greater range of foods and to expand their range into darker and colder places. Just when and how early human ancestors began using fire is debated and may never be precisely known. It is likely that early fire was "captured" from the wild, e.g. from fires started

Fire may also have been used as a hunting tool by setting fire to forests to drive out game. Fire could also be used for protection to drive off predators.

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Naturally set fires were the most likely source of early fire. Burning embers may have been carried back to home sites, or the fire used where it was.

by natural events such as lightning. Whatever its origin, the controlled use of fire changed the course of both physical and cultural evolution, influencing dietary range and improving survival. Possible evidence exists for the use of fire up to a million years ago, but real evidence of controlled use of fire dates back only a few hundred thousand years.

The development of the hand-drill meant early humans could make fire when and where they needed it.

A not so obvious, but important use of fire, is its use in the construction of weapons. Wooden spear tips can be hardened in fire, allowing them to be made sharper and improving their penetrative power. Rocks heated in a fire become brittle and are therefore easier to shape into tools, such as blades and spearheads. The blade also holds a sharper edge.

Fire provides light and can be used as a torch, allowing early humans to explore dark places such as caves. It may have helped with bonding, such as sitting around a campfire. Fire provided light and warmth at night.

Cooking is the most obvious use of fire. Cooking food makes it easier to chew and digest, releasing energy more quickly for the body to use. Fire also kills parasites and pathogens in the food. Fire can also be used to preserve food. Meat can be smoked and vegetable material can be dried.

1. List four uses of fire:

2. (a) Describe two important consequences of cooking food:

(b) Explain how cooking food provided a selection pressure for a smaller jaw and teeth:

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3. Discuss the influence of fire on human cultural evolution:

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181 Shelter and Clothing

Key Idea: The development of clothing and techniques to build shelters allowed the expansion of humans into colder climatic areas. Human ancestors would have had a far greater amount of body hair than humans today. The move into hot open grassland provided a selection pressure for the reduction of

body hair. However this then limited the ability to survive in cold climates as humanity migrated north out of Africa into Europe and beyond. The use of animal skins or woven plant material to make clothing provided a way of keeping warm. Building shelters allowed humans to shelter from both the weather and possible predators.

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Early humans and their ancestors are often depicted in popular texts wearing animal skins. However there is little direct evidence of humans wearing clothes until 26,000 years ago (the evidence is a finely worked bone needle). Evidence from burial sites also shows humans were adorning themselves (or at least their dead) with necklaces as far back as 35,000 years ago. Evidence from the evolution of body lice suggests human ancestors began wearing clothes some 100,000 years ago. This finely worked bone needle is almost identical in shape to needles used today. This needle is dated between 17,000 and 10,000 years old.

Didier Descouens CC 3.0

Liang Bua cave, Flores Island, Indonesia.

Rosino

Although often called cavemen, human ancestors probably didn't start using caves regularly as shelter until they could control the use of fire. Bones of early australopithecines or Homo found in caves were likely taken there by predators or been washed in by ancient streams. Some of the first evidence of caves being used by humans is from the Atapuerca cave in Spain. It is dated at 1.2 million years old, but it is difficult to tell the exact relationship of the fossils to the cave.

Hide drying on rack

The first evidence of humans building purpose-built shelters dates to around 400,000 years ago. Post holes in the ground indicate poles were used as scaffolding for the shelter. Preserved bones from around 17,000 years ago in the Ukraine show humans there used mammoth bones and tusks as scaffolding, probably because there would have been few trees available to make wood scaffolding from.

1. (a) Identify the evidence for humans building purpose built shelters:

(b) Identify the evidence for humans making clothing:

(c) Suggest a reason why there is so little evidence of humans building shelters or making clothes:

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3. What is the main purpose of clothing?

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2. Suggest a reason why caves where probably not regularly used by humans and there ancestors until the controlled use of fire was developed:

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182 Art and Spirituality western and eastern Europe. Growing evidence suggests Neanderthals too were culturally sophisticated. The stimulus for the new cultural development was probably a need to represent ideas about the unknown, such as death, hunting success, and fertility, in a concrete way. A wide range of materials were used to do this. Ivory, bone, clay, and stone were used to create sculptures, and the walls of rock shelters and caves were adorned with drawings, paintings, and bassrelief (sculptures that stand out from the rock wall).

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Key Idea: Human art and culture can be dated to as far back as 290,000 years. However a major cultural explosion took place about 35,000 years ago. Until recently, it was believed that art and spiritual beliefs first developed with the arrival of modern humans, particularly in Europe. However ancient artworks dating to nearly 300,000 years ago can be found outside of Europe. The beginning of a period 35,000 years ago marks a dramatic cultural development occurring simultaneously over large parts of

Bronze age begins in Europe 3200 BC

0 AD

Iron age begins in Europe 1500 BC

Neolithic era begins in Europe 4000 BC

Earliest example of pottery. Found in Xianrendong Cave. Dated to 18,000 BC.

Earliest Japanese art. 14,000 BC

25,000 BC. Earliest ceramic art.

25,000 BC

Earliest of the European Venus figurines (37,000 BC).

The earliest Asian art is dated to 38,000 BC, found in the Sulawesi Cave in Indonesia.

Oldest known parietal art (cave paintings) Cave of El Castillo, 39,000 BC.

Earliest musical instruments. There is debate over the dating of various flutes but the oldest are at least 43,000 years old.

50,000 BC

Neanderthal artists create the La Ferrassie Cave Cupules in France, dated to 60,000 BC.

JosĂŠ-Manuel Benito CC 2.5

Earliest African art, the Venus of Tan-Tan, dated to 100,000 BC.

100,000 BC

200,000 BC

Bhimbetka petroglyphs, India, dated to at least 290,000 BC. Consist of depressions (cupules) hammered into the rock of the Bhimbetka cave.

Most Venus figurines (right) are small (approx 10 cm high) figurines of women with exaggerated breasts, buttocks, and body fat. They may have represented desirable traits among women to enhance fertility and survival. Venus of Berekhat, dated to 230,000 BC. The oldest known mobiliary art (figurines).

Don Hitchcock CC 3.0

This Aurignacian (Upper Palaeolithic) flute, made from an animal bone, is about 43,000. A similar flutelike piece of cave bear bone has been found at a Neanderthal hunting camp in Slovenia. The bone, also dated at about 43,000 years ago, suggests that Neanderthals may have made music but there is debate over the dating.

Ubirr rock paintings. Earliest known art in Australia (and Oceania) dated to 30,000 BC.

Venus of Willendor (30,000 years old)

300,000 BC

1. The earliest artworks discovered in Japan include small but elaborate figurines called dogu. Why are the earliest artworks in Japan so elaborate, whereas the earliest artwork in Europe or Africa is often very simple?

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3. What might the various Venus figurines have represented?

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2. The earliest examples of musical instruments are about 43,000 years old. Explain why it is likely musical instruments were probably used well before this date:

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183 Mesolithic and Neolithic Cultures climate would have resulted in increased productivity. The date of the Neolithic (New Stone Age) culture varies with geographic location. In the middle East it dates from about 10,000 BC, and in Europe from 4000 BC. In parts it was short and replaced by bronze technology relatively quickly. Domestication of animals and the development of farming allowed a shift away from the hunter-gatherer economy of the Mesolithic to a food producing culture. Not all individuals had to be involved in food gathering activities, and some people developed specialised craft skills (e.g. potters). As permanent settlements developed, ideas and knowledge could be more easily transferred between people, resulting in a rapid expansion of cultural evolution during this period.

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Key Idea: The Mesolithic is characterised by refinement and further development of human tools. The Neolithic is characterised by advances in farming and animal domestication practices, the development of crafts (e.g. pottery and weaving), the use of polished stone and flint tools, and the development of permanent settlements. The Mesolithic (Middle Stone Age) period occurred in Europe from about 10,000 to 4000 BC, as the last glacial period ended. The tools produced at the time were small bladed geometric stone tools (microliths), and were often fitted into a handle of wood or bone. Mesolithic people used a wide variety of hunting, fishing, and food gathering techniques, which may have arisen because the warming

The origin of agriculture

Mesoamerica

Fertile Crescent

Beans, maize, peppers, squash, gourds, cotton, guinea-pigs, llamas

Barley, wheat, Emmer, Einkorn, lentils, peas, sheep, goats, cattle

5000 years ago

10,000 years ago

North China

Southest Asia

Rice and millet

Rice, bananas, sugar cane, citrus fruits, coconuts, soya beans, yam, millet, tea, taro, pigs

7000 years ago

Maize

South America

Lima beans, potatoes, squash, beans, and pumpkins

Africa

Millets, sorghum, groundnuts, yams, dates, coffee, and melons

Each domesticated animal was bred from the wild ancestor. The date indicates the earliest record of the domesticated form.

Domesticated animal

Wild ancestor

Region of origin

Date (years ago)

Domesticated animal

Wild ancestor

Region of origin

Date (years ago)

Dog

Wolf

many places?

13,000

Horse

Wild horse

Southern Ukraine

6000

Goat

Bezoar goat

Iraq

10,000

Arabian camel

Wild camel

Southern Arabia

5000

10,000

Bactrian camel

Wild camel

Iran

4500

Sheep

Asiatic mouflon Iran, Iraq, Levant Aurochs

Southwest Asia

10,000

Llama

Guanaco

Andean plateau

6000

Pig

Boar

Anatolia

9000

Water buffalo

Indian wild buffalo

Indus Valley

4500

Domestic fowl

Red jungle fowl

Indus Valley

4000

Ass

Wild ass

Northeast Africa

5500

(a) State when the Mesolithic culture began:

(b) Explain how it differed from the Upper Palaeolithic culture:

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1. The Mesolithic culture replaced the Upper Palaeolithic culture.

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Cattle

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Mesolithic and Neolithic tools Sickles: These two sickles are made of flint embedded into a handle made of horn (below, right) and an antler (right). Such tools were used to cut the grasses to gather their seeds and date from the Mesolithic period.

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Grindstone: This Neolithic grindstone was used to grind grain so it could be used in cooking.

Antler handle

Single flint blade

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JosĂŠ-Manuel Benito

Three flint microliths embedded to create cutting edge

Bone fish hook: This fish hook dates from the Mesolithic period and was found in Sweden.

Sandstein

Microlith: Made of flint or chert. Microliths formed the points of hunting weapons such as spears. This microlith was found in the Tourasse cave, France.

Didier Descouens

Horn handle

Neolithic people produced a wide range of purposespecific tools to harvest, store, and prepare food. The photo, above, displays food and cooking items retrieved from a Neolithic site in Europe. The containers are made of antlers and wood.

The development of agricultural practices significantly changed the lifestyle of Neolithic people. The hunter-gatherer lifestyle gave way to food production using agricultural techniques. The need to tend to crops and animals resulted in a shift away from a nomadic lifestyle to living in permanent settlements. Skara Brae, Scotland, is the location of Europe's most complete stone-built Neolithic village (above). It consisted of ten houses, and was occupied between 3180 and 2500 years ago. The people of Skara Brae raised cattle and sheep, and may have cultivated barley. The presence of fish bones and shells amongst the ruins suggest their diet was supplemented with seafood from the nearby ocean.

Neolithic village at Skara Brae, Scotland.

(a) State when the Neolithic culture began:

(b) Describe the important cultural developments of the Neolithic culture:

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2. The Neolithic culture replaced the Mesolithic culture.

3. Why was the shift of Neolithic people from hunter-gatherers to farmers an important step in cultural evolution?

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Dr J.F Burka: http://en.wikipedia.org/wiki/File:Orkney_Skara_Brae.jpg

The development of permanent settlements


184 Hominin Evolution: Probable Phylogenies

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Key Idea: Homo sapiens are the last representative of a once extensive, multi-branched hominin evolutionary tree. Distinguishing the human lineage from all these species can be difficult and in some cases open to interpretation. The diagram below shows a possible evolutionary history of hominins, demonstrating the fact that human evolution was not a linear sequence and that many phylogenies are tenable. There is much controversy over the interpretation

of fossil data. The hominin lineage underwent an adaptive radiation about 3 mya, producing many different species. The genus Australopithecus gave rise to the genus Homo and the genus Paranthropus, which coexisted with early Homo, but eventually became extinct about 1 mya. The genus Homo is represented by many species as successive waves migrated out of Africa. Homo sapiens, which migrated out of Africa 80,000-60,000 years ago, is now the only living species.

*The species marked with an asterisk (*) were all unknown a decade or so ago. There are likely to be many as yet ‘undiscovered’ species in the fossil record between 7 and 2 million years ago.

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Denisovan hominin dated at > 30,000 (Russia). Species currently not assigned.

0

Homo sapiens

Homo neanderthalensis

Homo naledi was discovered in 2013 in Rising Star Cave, South Africa. Initially difficult to date, in 2016 techniques to date fossil teeth, surrounding sediments, and overlying flowstone provided a date of 335,000-236,000 years.

Homo Homo floresiensis erectus

*

?

*

?

Homo naledi

Homo heidelbergensis

*

Homo antecessor

1

*

Homo habilis

Homo ergaster

?

Homo georgicus

*

2

Time (millions of years ago)

Homo rudolfensis

Paranthropus robustus

Australopithecus sediba

Paranthropus boisei

*

Australopithecus africanus

? ?

Australopithecus garhi

Finds such as H. georgicus generate debate over the various models of the hominin lineage. H. georgicus is found outside Africa somewhat earlier than expected.

?

Australopithecus afarensis

?

*

Paranthropus aethiopicus

?

3

Kenyanthropus platyops

?

?

Australopithecus deyiremeda

Australopithecus bahrelghazali

?

*

Australopithecus anamensis

*

4

?

*

?

Species in bold are representative of the trends seen in hominin evolution.

Ardipithecus ramidus ramidus

*

5

6

? Sahelanthropus tchadensis

*

7

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*

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Orrorin tugenensis

Ardipithecus ramidus kadabba

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Group

Examples

Homo sapiens, Homo neanderthalensis, Homo heidelbergensis

There have been many recent discoveries of relatively recent Homo species throughout Eastern Europe and Asia. How these fit into the hominin lineage and how they are related to modern humans are still debated. These specimens influence the interpretations of early human migration across the world.

Homo erectus, Homo ergaster, Homo floresiensis

Homo erectus and closely related species show increasingly sophisticated tool cultures. H. erectus spread throughout Asia. There is much debate over how the different erectine species are related, with some paleontologists suggesting that all the erectines should be labelled as one species.

Homo habilis

This group shows the first signs of brain enlargement, more meat in the diet as well as the first recognisable stone tool culture. The post-cranial (below the head) skeleton remains small and slight, much like that of the australopithecines.

Paranthropines

Paranthropus robustus, Paranthropus boisei

These early hominins represent a group specialised for eating a bulky, low-grade vegetarian diet. They evolved large cheek teeth, powerful chewing muscles and a generally robust skull (large crests for muscle attachment, heavily buttressed face).

Australopithecines

Australopithecus afarensis, Australopithecus africanus, Australopithecus sediba

The earliest australopithecines were among the first apes to achieve bipedalism. They possessed a gracile body form and were probably opportunistic omnivores, scavenging meat from carcasses and exploiting a range of resources.

Ardipithecus ramidus, Orrorin tugenensis

Essentially chimpanzee-like animals that have begun to show some human characteristics in their locomotion (bipedalism) and in the shape and arrangement of their teeth. Sahelanthropus tchadensis may be a common ancestor of chimpanzees and humans.

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Archaic and modern humans

Erectines

Habilines

Very early hominids

1. On the opposite page, colour in the blue outlined species boxes so that all species in the group are coloured the same: archaic and modern humans (orange), erectines (green), habilines (red), paranthropines (yellow), australopithecines (black), and very early hominins (blue). 2. The relationships between ancestral hominins is open to interpretation. The diagram opposite allows for various phylogenies to be constructed as hypotheses because the exact lineages are unknown. Starting at Australopithecus afarensis, complete three likely lineages for the evolution of Homo sapiens by writing the species in order of appearance. You can use the dotted lines to help you but other paths may be possible. The first one has been started for you:

Australopithecus (a)

afarensis , Homo habilis, Homo ergaster...

(b)

(c)

5. Which hominin is the probable most recent ancestor of Homo sapiens? Š2020 BIOZONE International ISBN: 978-1-98-856647-4 Photocopying Prohibited

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4. Which hominin existed for the longest length of time?

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3. Explain how finds such as Homo naledi and Homo georgicus force us to rethink our understanding of how and where humans evolved:


185 The Importance of Ardi

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Key Idea: Analysis of the skeleton of Ardipithecus ramidus has produced findings that suggest bipedalism and a manipulative hand are very ancient features. The first fossils of Ardipithecus ramidus were discovered in the Middle Awash region of northeastern Ethiopia in 1994. After many years of excavation, a partial skeleton was unearthed. Studying the skeleton of Ar. ramidus is beginning to change

our understanding of hominin evolution. Until recently, it had been theorised that our earliest ancestors moved about very much like the chimpanzees of today. However the evidence from the Ar. ramidus skeleton shows that this is not the case and that bipedalism developed in quite a different way to what was once thought. Moreover, a dextrous hand developed early and is also an ancient trait.

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Brow ridge

Skull: The skull of Ar. ramidus shares certain features with Australopithecus, including a reduction in the size of the canine teeth in both male and females. This implies a reduction in aggression between males. The orientation of the base of the skull on which the brain stem rests suggests that the parts of the brain involved in visual and spatial perception were already beginning to develop.

Reduced canine teeth

Ardipithecus ramidus skull features

Lack of strengthened knuckles

The primitive features of the Ardipithecus hand that are shared with Homo.

Human hand

Features associated with bipedalism that are shared by Ardipithecus and Homo

Socket joint for femur

Feature associated with tree climbing that is shared by Ardipithecus and Pan

Foot: The foot of Ar. ramidus is a generalised one, with some human-like features, such as a rigid foot, as well as some modern ape-like features, such as an opposable big toe. These features indicate that Ar. ramidus spent considerable time climbing in trees.

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Well developed thumb

Short, flexible palm

Human pelvis

Images redrawn from C. Owen Lovejoy, et al Science, vol 326, 2009

Short metacarpals

Ardipithecus ramidus pelvis

Chimpanzee pelvis

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Pelvis: The pelvis of Ar. ramidus indicates that the modern pelvis, evolved for bipedal locomotion, began its evolution in the trees. Although several features of the upper pelvis strongly indicate bipedalism, features of the lower pelvis show that muscles associated with tree climbing were still well developed.

Prognathic muzzle

The foot is rigid in both Ardipithecus and Homo

Big toe points to the side Homo sapiens foot

Big toe points forward

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Hand: Analysis of the hand of Ar. ramidus shows that it is similar to our own, and that human hands are therefore close to the primitive form and not as greatly modified for tool use as was previously thought. Ar. ramidus had a flexible wrist and the opposable thumb was well developed. By contrast, chimpanzees move on the ground by knuckle walking, a motion that requires strengthening of the wrist and knuckle bones and lengthening of the palm, making the hand less flexible and not as dextrous.

Relatively small cheek bones

Ardipithecus ramidus foot

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Proposed evolutionary tree with Ardipithecus Pan (chimpanzee) - Knuckle walker - Skilled climber - Short stiff back - Flexible, grasping feet - Strengthened wrist and elongated palm - Large incisors for eating fruit - Similar size in males and females - Canine teeth in males larger than in females

Homo - Habitual upright walker - Terrestrial biped - 'S' shaped flexible lower back - Multiple environment omnivore - Similar size in males and females - Small canines in males

Ardipithecus - Facultative upright walker - Able tree climber - Retained long flexible lower back - Woodland and forest omnivore - Similar body size in both sexes - Small canine teeth in males

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Modified from C Owen Lovejoy, et al Science, vol 326, 2009

Gorilla

Chimpanzee-human Last Common Ancestor (CLCA) - Long, flexible lower back - Short flexible palm and wrist - Palm walking tree climber (not brachiating or knuckle walking) - Canines larger in males

This simplified evolutionary tree of hominids shows that chimpanzees have continued to evolve into a specialised tree climber and are not simple modifications of the Chimpanzee-human Last Common Ancestor (CLCA). Indeed, the CLCA was an ape-like creature with many generalised features that have undergone further modification in both humans and chimps alike.

1. Describe the evidence for reduced aggression between Ar. ramidus males:

2. Explain why the human hand might now be viewed as the primitive type:

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3. Describe the evidence for bipedalism developing in a primarily arboreal (tree-dwelling) ancestor rather than in a knuckle walking, terrestrial ancestor:

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4. How does having a generalised body plan increase possible evolutionary pathways?


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186 New Findings: Denisovans

Key Idea: DNA evidence suggests that early modern humans interbred with a previously unknown Homo species. The finding of fragments of bone and teeth in a cave in Siberia indicated that another, until then unknown Homo species once

lived in Asia. The Denisovans, (after the cave in which the fossils were found) have yet to be assigned a species name. They are sometimes called Homo sapiens ssp. Denisova.

The Denisova cave finds

ffIn 2008, archeologists discovered a fragment of finger

bone in the Denisova cave, in Siberia. The bone fragment belonged to a juvenile female (named X-woman).

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ffArtefacts, such as a bracelet, were found at the same level as the finger bone.

ffIn 2010, a molar tooth was found at a different level to the

finger bone, indicating it belonged to a different individual. A toe bone found in 2011 was at the same level as the tooth.

ffThe molar found in the Denisova cave has unique

characteristics, which are not present in the molars of Neanderthals or modern humans.

ffCarbon dating estimates the age of the artefacts and bone fragment at 40,000 years.

temperature and in acidic soil conditions. The cool temperatures within the Denisova cave preserved the DNA in the fossil fragments. The fossils contained very low levels of DNA contamination from other organisms.

The Denisova cave, in the Altai mountains, Siberia, Russia

Obersachse

ffFossil DNA degrades quite rapidly with increasing

Using genome analysis to classify the Denisova cave fossils African French

Han Chinese

Possible interbreeding ~100,000 years ago

Interbreeding ~60,000 years ago

Melanesian

Interbreeding ~40,000 years ago

Denisova

?

~400,000 years ago

Vindija Neanderthal

~300,000 years ago

Source: New Scientist 13 Aug 2011

Nuclear DNA analysis suggests the Denisova fossils belong to a previously unknown hominin species that existed at the same time as modern humans and Neanderthals, but was genetically distinct from them (above). The fossils are called the Denisovans, because they have not yet been formally classified.

Nuclear genome analysis suggests the Denisovans were a sister group to the Neanderthals. They probably shared a more recent common ancestor with Neanderthals (~300,000 years ago) than with present day humans (~400,000 years ago).

1. Why are the Denisovans difficult to classify?

2. (a) What modern human lineage appears to have interbred with the Denisovans:

(b) What percentage of DNA does this lineage appear to share with the Denisovans?

3. Why was the Denisovan DNA in remarkably good condition?

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A Melanesian woman

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The Denisovan's interbred with the ancestors of the present day Melanesians (right), and possibly with the Neanderthals, but not the ancestors of other present day populations, such as the Han Chinese. Melanesian DNA includes between 4% and 6% Denisovan DNA.

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187 New Interpretations: The Neanderthals closest relative to modern humans, so there is considerable interest in mapping the Neanderthal genome. By comparing the Neanderthal genome to the genome of present-day humans, it may be possible to identify genes in modern humans that have been influenced by positive selection.

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Key Idea: New, more complete analysis of Neanderthal DNA is revealing multiple interbreeding events with early humans. Neanderthals appeared about 400,000 years ago, and disappeared 25,000-30,000 years ago. They lived in Europe and parts of western and central Asia. Neanderthals are the

Front view Neanderthal skull

La Ferrassie Neanderthal skull, France

Image: Bone clones

Difficulties in analysing Neanderthal DNA ffThe DNA is often degraded to small fragments less than 200 base pairs

long. This makes it difficult to obtain sequence overlaps (critical for assembly of the genome).

ffThe DNA is often of poor quality because it has been chemically modified and degraded by the environment.

DNA being extracted from a Neanderthal fossil

95-99% of the DNA obtained from the Neanderthal fossils analysed was from microbes that colonised the bone after the Neanderthal died. Researchers must be careful not to contaminate the sample with their own DNA.

What has been found?

The Neanderthal legacy

The continuing analysis of Neanderthal (and Denisovan) DNA has found that there were at least five interbreeding events between humans (H. sapiens), Neanderthals, and Denisovans.

ffBetween humans and Neanderthals. Analysis shows that

Analysis of Neanderthal DNA published in 2016 suggests that they carried various mutations that made them up to 40% less reproductively fit than modern humans. When interbreeding occurred with humans, some of these mutations would have been passed to the human gene pool. Over time most of the harmful mutations were discarded through natural selection, but some have remained. Other genes that may have been beneficial also entered the gene pool. However the benefits that these genes once conferred may no longer exist as the human lifestyle becomes more sedentary and diets change.

ffSome human populations that migrated east across Eurasia

Studies matching health problems to Neanderthal DNA have found that genetic variants inherited from Neanderthals are linked to an increase in the risk of heart attacks, depression, skin disorders, and nicotine addiction. However, the Neanderthal DNA may not necessarily be causing the health problem. It might just be associated with human DNA that is.

ffBetween archaic humans and Neanderthals. Analysis in

2016 shows there may have been an interbreeding event around 100,000 years ago when an early wave of humans migrating out of Africa met a group of Neanderthals migrating from Europe to Asia. between 1-4% of the genomes of people outside of Africa is derived from Neanderthals (more than for Africans). It is thought these encounters may have occurred as humans migrated out of Africa around 50,000 - 60,000 years ago and met Neanderthal populations already in the Middle East. interbred with the Denisovans. Evidence of this in found in Melanesian DNA (see opposite).

ffDenisovans also interbred with Neanderthals, probably about 50,000 years ago (see opposite).

ffDenisovans interbred with an unknown group of hominins,

possibly an offshoot of H. erectus, about 100,000 years ago.

Some genes that were possibly inherited from Neanderthals or Denisovans have provided benefits. Tibetans appear to have inherited Denisovan genes that enabled high altitude adaptation. Humans may also have inherited genes associated with immunity to new diseases found outside of Africa, but already encountered by Neanderthals.

2. In which group of modern humans is Neanderthal DNA mostly found and why?

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1. What percentage of Neanderthal DNA is present in modern humans?

3. Describe some possible positive and negative effects of Neanderthal DNA in modern humans:

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NIH

ffSamples are often contaminated with the DNA of other organisms. Between

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188 Problems With Interpretation: H. floresiensis

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Key Idea: The discovery of Homo floresiensis has proved highly controversial with several hypotheses proposed to explain its position in hominin evolution. In 2003, hominin fossils, including an almost complete skeleton, were discovered on the island of Flores, Indonesia. The fossils were assigned to a new species, Homo

floresiensis, thought to have lived on the island as recently as 18,000 years ago. However, revised dates in 2016 indicate that H. floresiensis lived ~190,000-50,000 years ago. The new date is close to the time that modern humans reached the area, suggesting that encounter with H. sapiens may have contributed to the demise of the Flores population.

The fossils were discovered in Liang Bua, a limestone cave on Flores Island, Indonesia. The cave contains 12 m of stratified deposits. The remains of modern humans, as well as Homo floresiensis, have been found in the cave.

Rosino

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Homo floresiensis was very small and fully bipedal. Although the brain was very small, its organisation was advanced and stone tools associated with the skeleton indicated well established hunting technology. In contrast to these features, aspects of the skeleton showed primitive features found only in apes and early hominins. H. floresiensis probably lived 190,000 to 50,000 years ago, making it one of the latest surviving hominins along with the Neanderthals and the Densiovans. Its discovery caused widespread controversy and several opposing hypotheses were put forward to explain its place in human evolution. Since its discovery, earlier fossils dating to 700,000 years old have been found.

Brain size: Very small, only 380 cc. 3D scans showed expansion of the prefrontal cortex and temporal lobes, brain regions associated with complex thought.

Different interpretations of the fossils

The discovery of Homo floresiensis caused a storm of controversy, not helped by the fact that the specimens were removed from their repository by one palaeontologist, kept from other scientists for three months, and returned damaged. The Indonesian government also denied scientists access to the cave where the fossils were found for two years. Several hypotheses were eventually put forward to explain how such a small statured hominin evolved or survived. These included:

Facial features: Relatively modern dentition, but teeth are large relative to the rest of the skull.

ffH. floresiensis evolved a result of island dwarfism. Island

dwarfism is a relatively common occurrence where large animals become smaller over time when isolated, e.g. the extinct pygmy elephants on Flores showed this adaptation.

No chin present.

ffH. floresiensis is not a new species but instead an individual with a disease or disorder. Some scientists put forward the idea that the small skull was a result of microcephaly, a neurodevelopment disorder. Others thought that perhaps H. floresiensis suffered from congenital hypothyroidism.

ffDetailed examination of the bones showed no overlap with

Sc

CC na

y ail

4.0

any features expected from individuals with the diseases or disorders listed above. A study of the bones and joints of the arm, shoulder, and lower limbs concluded that H. floresiensis was more similar to early humans and apes than modern humans. Small brain size coupled with more advanced brain organisation indicate possible parallel evolution of sapiens-like features.

Homo floresiensis

Homo sapiens

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2. Describe the different interpretations of the H. floresiensis bones:

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1. Explain the effect that the Flores finds had on the hypothesis that hominins continually evolved larger brains and bodies:

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189 Problems with Dating: H. naledi dozens of teeth. Interestingly, no other types of animal or plant have been found in the cave and there is no indication of water flowing in the past. This has led the investigators to hypothesise that the bodies were deliberately placed there by other Homo naledi. Also the lack of other flora or fauna and sediments (other than cave dust) has made it very difficult to date the fossils. They have both advanced and primitive features, which some think puts them at the cusp of the transition between Australopithecus and Homo.

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Key Idea: The discovery of Homo naledi presents problems. The placement of the bones appears deliberate, which seems unlikely for an ancient hominin, and they are difficult to date because there are no dateable sediments around them. In 2013 two cavers exploring the Rising Star cave system near Johannesburg in South Africa found a passage that led to a chamber containing the bones of a new hominin species. Excavation has found more than 1500 specimens from at least 15 individuals including ribs, skulls, jaws, and

The Rising Star cave system

Cave entrance

Superman's crawl may have been higher in the past, allowing easier access.

Bodies were probably dropped down the shaft over a long period of time (maybe centuries)

Dinaledi chamber

Dragon’s back

10 metres

Superman’s crawl (less than ten inches high)

Fossil find

Dating Homo naledi ffDating the Homo naledi bones is problematic because they

were found deep in the cave. Ordinarily, fossils can be dated by relative dating. Other fossils in the sediment can be used to date the unknown fossils. For example, if the bones of a predator species of a known age had been found in the cave, then it may imply that H. naledi lived at the same time. However, only H. naledi bones have been found, expect for a few small birds on the surface. Radiocarbon dating cannot be used because it only dates accurately to 50,000 years of age and H. naledi is likely to be much older than that.

ffIf the bones had been washed into the cave by a river system

it may have been possible to use the sediments deposited or other bones that had been washed in to provide a date. The excavation team has yet to find any evidence of a river or water flow. One other way of dating the bones is by dating the flowstones found in the cave. Flowstones are sheets of calcium carbonate built up by water flowing down cave walls (similar to stalactites). However the flowstones do not cover much of the cave floor and fossils. providing a date of ~1-3 my. However, more recent radiometric dating of specimens and the overlying flowstones has provided a much more recent date of 335,000- 236,000 years.

Lee R Berger CC 4.0

ffVarious anatomical ways of dating the fossils were tried, initially

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2. Describe two reasons why dating the H. naledi bones is difficult:

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1. Describe one of the problems in explaining the deliberate placement Homo naldei bones in the Dinaledi chamber.

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190 The Origin and Dispersal of Modern Humans

Key Idea: Hominins evolved in Africa. Waves of migration of Homo sapiens out of Africa saw anatomically modern humans eventually reach all parts of the world. The map below shows a suggested probable origin and dispersal of modern humans throughout the Old World into the Americas. An African origin is almost certain, with south eastern Africa a likely region. The dispersal was affected at crucial stages by the last glacial, when ice sheets covered

much of Europe. Dispersal was also affected by the presence or absence of land bridges formed during the drop in sea level associated with the onset of glacials. Recent evidence suggests that island-hopping and coastal migration may also have been important, e.g. for the movement of people into Indonesia. The late development of boating and rafting technology slowed dispersal into Australia and the Pacific. On the map, ya = years ago.)

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KEY

The first modern humans appear in Europe 45,000 years ago

Over land migration route

Coastal migration / island hopping

Earliest modern humans travel across the temporary ice-age land bridge to the Americas between 30,000 and 15,000 years ago

Cro-Magnon 30 000 ya

Qafzeh 120,000 - 92,000 ya

Skuhl 101,000 - 81,000 ya

The earliest dating in East Asia for modern humans is 80,000 years ago from Daoxian in southern China

Morocco 315,000 ya

Afar, Ethiopia 160,000 ya Omo 130,000 ya

Recent finds from Morocco suggest that AMH emergence may have involved the whole of the African continent.

Polynesia populated progressively between 4500 - 700 years ago

Malakunanja II 50,000 ya

Border Cave 115,000 - 62,000 ya

Australia was first occupied by modern humans at least 50,000 years ago.

Lake Mungo 31,000 ya

Klasies River Mouth 120,000 - 84,000 ya

New evidence suggests first n permanent Maori settlement in New Zealand only 700 years ago (not 1200 years)

One origin or many?

Two primary hypotheses have been put forward to explain where modern humans evolved and what happened to the Homo species that preceded them.

Replacement model

In this model, modern Homo sapiens emerged gradually throughout the world and, as the populations dispersed, they remained in 'genetic contact'. This gene flow between neighbouring populations ensured that the general 'modern human blueprint' was adopted by all. Limited gene flow still allowed for slight anatomical differences to be retained or develop in the regional populations. This model is based largely on fossil evidence and the anatomical characteristics of modern populations, but advocates of the model maintain that the mitochondrial DNA data can be interpreted in a way that supports it.

Also known as 'Out of Africa' and the 'Eve Hypothesis'. This model states that modern humans evolved from archaics in one location, Africa, and then spread, replacing the archaic populations, without interbreeding. Modern human variation is thus a relatively recent phenomenon. Mitochondrial DNA (mtDNA) analysis of modern endemic human populations showed that the highest level of genetic variation in mtDNA occurs in African populations, implying that H. sapiens arose first in Africa. Using the genetic distance between African populations and others as a measure of time, this model places the origin of modern humans at back to a single female (Eve) who lived there some 200,000 years ago.

(b) At about what date did this migration occur?

2. How many successive waves of Homo migrated out of Africa?

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1. (a) Which hominin was the first to migrate out of Africa?

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Multiregional model

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Evidence for dispersal There is little doubt that humans evolved in Africa. Fossils of ancient hominins such as Australopithecus afarensis, A. africanus and Homo habilis are all found in Africa. Analysis of fossils and genetic evidence supports theories that Homo erectus migrated out of Africa about 1.8 mya. Populations of H. erectus remaining in Africa gave rise to H. heidelbergensis, which also migrated out of Africa about 650,000 years ago but remained in Europe and Western Asia. Those populations gave rise to H. neanderthalensis. Homo heidelbergensis in Africa eventually gave rise to H. sapiens. H sapiens began to migrate out of Africa between 120,000 and 80,000 years ago but the movement was into Europe was halted by a period of cold climatic conditions. Around 60,000 years ago the final migration of modern humans out of Africa and across Europe and Asia began. Analysis of various loci in the human genome show humans have very little genetic diversity. The analysis shows that outside of Africa, human genetic diversity is a sub-set of African genetic diversity.

S

China

Africa N E

W+S Asia

M. East

MYA 3.0

Europe

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Acheulian stone tools are found throughout Africa, Europe, and South and Western Asia, but appear to be largely absent from Eastern Asia. Tool cultures develop first in Africa then move to other parts of the world.

1.0

Diversity (0 = low)

Locus (measurement)

Africa

Asia

Europe

30 microsatellites

0.807

0.685

0.730

Xq 13.3 (short arm X chromosome)

0.035

0.025

0.034

50 autosomal sequences

0.115

0.061

0.064

mt DNA control region

2.08

1.75

1.08

0.3

0.1

0.03

0.01

Oldowan Acheulian/ Palaeolithic

Acheulean-like Neolithic

Diversity in the CD4 locus on chromosome 12. Blue shading represents the amount of diversity.

3. Evaluate the evidence in this activity and decide which model (Multiregional or Out of Africa) of human dispersal is more likely and explain why:

4. When did modern humans appear in Europe?

5. How did humans manage to migrate into the America?

(b) Suggest why New Zealand was one of the last land masses populated by humans:

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6. (a) Study the map above and suggest why humans reached Australia before they reached Western Europe, even though Australia is further away from their point of origin:


191 A Summary of Trends in Hominin Evolution

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274

Key Idea: Specific trends can be seen in human physical (biological) and cultural evolution. Use this activity to revise your knowledge about trends in human evolution. Cut out the images on page 275 and place them in their correct place on the timeline of human evolution (below and on page 275). Add notes about significant

Millions of years ago

developments and trends. Notes may include information about the tool technology (including the main user of the tool), skull features and brain size (represented by blue circles) bipedalism, how food and diet influenced dentition and hominin distribution.

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3

2

Skull and body features

Jaw shape

LINK

REVISE

LINK

LINK

LINK

LINK

172 173 174 175 179

CL

Brain size and features

N AS OT SR F OO OR M US E

Tools

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1350 cc Homo neanderthalensis

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Australopithecus afarensis

12001700 cc

Homo habilis

450 cc

Homo erectus

Archaic Homo sapiens

500800 cc

10001250 cc

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375550 cc

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Homo sapiens


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This page has been deliberately left blank

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0

N AS OT SR F OO OR M US E

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2


192 KEY TERMS AND IDEAS: Did You Get It?

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278

1. Test your vocabulary for this chapter by matching each term to its correct definition.

A Possessing a slighter frame. Usually used when referring the skeletons of

bipedal

A. afarensis and A. africanus, in contrast the to robust skeletons of the closely related Paranthropus.

carrying angle

B The group consisting of modern humans and their extinct ancestors. C The group commonly known as the apes. Examples include gibbons, chimpanzees,

gracile

PR E O V N IE LY W

and humans.

hominins

D Literally meaning “old stone” refers to the time of the stone tool cultures that lasted

hominoids

E Modern humans. Appeared around 160,000 years ago. Features include a large

from 2.5 million to 12,000 years ago. domed skull and upright forehead.

Homo

F Standing upright and walking on two feet.

H. neanderthalensis

G The forward projection of the face and lower jaw.

H Homo species that may have split from the lineage that led to H. sapiens around

600,000 years ago. Skeleton was more robust than modern humans and the skull was slightly larger.

H. sapiens paleolithic

I Genus of hominin of which only H. sapiens is still extant. Features include a fully

prognathic

J The carrying angle of the femur to the knee, which in humans and their ancestors brings the knee under the centre of gravity.

bipedal gait and large cranial capacity.

2. Draw lines to match up the hominid name with its meaning, then match the name to the correct skull number (1-8 below). Skull number:

Name:

Meaning:

Australopithecus afarensis

Man from the Neander Valley.

Australopithecus africanus

Handy man

Homo neanderthalensis

Southern Ape from Afar, Ethiopia

Homo sapiens

Southern Ape from Africa

Homo habilis

Man from the island of Flores or Flores man

Homo erectus

Knowing, or wise man

Homo floresiensis

Upright man

5

TEST

4

3

6

7

N AS OT SR F OO OR M US E

2

CL

1

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193 Review: Unit 4 Area of Study 1

Summarise what you know about this topic under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts. Use the images and hints to help you and refer back to the introduction to check the points covered:

Changes in biodiversity

HINT: Evidence for evolution. Dating fossils. Patterns of evolution.

Changes in genetic makeup of a population

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HINT: Mutations produce variation. Natural selection acts on phenotypes. Selective breeding.

REVISE


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280 Determining relatedness between species

Human change over time

HINT: Trends on physical and cultural development.

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N AS OT SR F OO OR M US E

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HINT: Comparing proteins sequences. Molecular clocks. Constructing phylogenies

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194 Synoptic Assessment: Unit 4 Area of Study 1

1. HIV/AIDS has killed more than 25 million people globally, and infected another 33 million since it was first recognised in 1981. In the mid 1990s, it was found that the HIV-1 virus entered T-cells of the immune system by docking with the receptor encoded by the CCR5 gene. Soon after this, it was discovered that the deletion of 32 bases in the gene (called CCR5D32) produces a premature stop codon in the mRNA and caused resistance to HIV-1.

% population with CCR5Δ32 mutation

18 16 14

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Geographical studies have found that the CCR5D32 mutation is found in light-skinned people of European descent in some areas of northern Europe where it is carried by up to 18% of the population. The mutation is virtually absent in Asian, Middle Eastern, and American Indian populations.

Western Europe has also been the site of many small pox epidemics before its elimination in 1980. Hypotheses for the prevalence of the CCR5D32 mutation vary. A possible hypothesis is that CCR5D32 may have conveyed some immunity to smallpox or plague or both.

12 10 8 6 4 2 0

Discuss the nature of the CCR5D32 mutation, including the significance of the premature stop codon in HIV resistance.

(a) What evolutionary pattern is shown by the Hawaiian fruit flies:

(b) Suggest why so many fruit fly species are present in Hawaii:

Oahu

Maui

Hawaii (youngest)

Photo: Karl Magnacca

The major dispersals of Drosophila and Scaptomyza.

(c) Describe the relationship between the age of the islands and the age of the fly species:

Kauai (older)

N AS OT SR F OO OR M US E

2. Drosophilidae (fruit flies) are a group of small flies found almost everywhere in the world. Two genera, Drosophila and Scaptomyza are found in the Hawaiian islands and between them there are more than 800 species present on a land area of just 16,500 km2. It is one of the densest concentrations of related species found anywhere. The flies range from 1.5 mm to 20 mm in length and display a startling range of wing forms and patterns, body shapes and colors, and head and leg shapes. Genetic analyses show that they are all related to a single species that may have arrived on the islands around 8 million years ago. Older species appear on the older islands and more recent species appear as one moves from the oldest to the newest islands.

(d) Account for this relationship:

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In 2004, a fossil of an unknown vertebrate was discovered in northern Canada and subsequently called Tiktaalik roseae. The Tiktaalik fossil was quite well preserved and many interesting features could be identified. These are shown on the photograph of the fossil below.

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282

The shoulder bones are not attached to the skull, allowing its neck to turn independently of the body.

Rod-like bones that help pump water over gills are present, but the presence of ribs indicates that lungs were also present.

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Tiktaalik's head is flattened horizontally like that of a crocodile with the eyes on top, looking up.

The bones of the limbs have a primitive pentadactyl arrangement, similar to tetrapods, which allowed it to support its body weight.

Fish-like fins are clearly visible.

do

he

og

ed

Gh

The fossil of Tiktaalik was covered with scales much like those of fish.

3. Use the information above to place Tiktaalik on the time line of vertebrate evolution. Discuss the evidence for your decision.

Jawless fish Bony fish Amphibians

Reptiles Birds

Mammals

150 mya

300 mya

365 mya

400 mya

550 mya

Vertebrate ancestor

4. The diagram below shows the foot of a chimpanzee, a human, and Australopithecus afarensis.

Chimpanzee foot

Australopithecus afarensis foot

N AS OT SR F OO OR M US E

Compare the foot structures above. Use the diagrams to analyse and describe the evidence that Australopithecus afarensis was bipedal. Discuss the selective advantages of being bipedal within the environment in which Australopithecus afarensis evolved. You may use more paper if required.

CL

Human foot

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DNA manipulation

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Unit 4 Outcome 2

Key terms

Enzymes enable us to manipulate DNA

annealing

Key knowledge

blunt end

CRISPR-Cas9 system

c

1

Describe how enzymes are used in the manipulation of DNA. Include reference to the principles and applications of endoncleases (restriction enzymes and the CRISPR-Cas9 system), ligases, and polymerases.

c

2

Explain how new gene editing technology allows genomes to be manipulated more easily, more precisely, and at less cost than ever before.

DNA amplification DNA ligase

DNA ligation

DNA polymerase

PCR amplifies DNA

gel electrophoresis

genetic modification

197

c

3

Describe DNA amplification using the polymerase chain reaction (PCR). Include reference to the role of primers and the thermally stable taq polymerase.

198

c

4

Describe when DNA amplification might be useful or necessary.

198

GMO

plasmid

196 - 197

Activity number

Key knowledge

gene technology

Activity number

polymerase chain reaction (PCR) primer

recognition site

recombinant DNA

recombinant plasmid restriction enzyme sticky end

taq polymerase

Gel electrophoresis sorts DNA fragments

vector

Key knowledge

Activity number

c

5

Explain the role of gel electrophoresis (of DNA) in DNA technologies. How does electrophoresis sort the fragments?

199

c

6

Explain how the DNA fragments on a gel are made visible and the role of DNA markers in identifying fragments of different size.

199

c

7

Interpret the results of gel runs. What sort of information can be provided by the banding patterns?

200

Recombinant plasmids can transform bacterial cells Key knowledge

Activity number 196

Explain what is meant by a recombinant plasmid. What genes are included in a recombinant plasmid?

c

9

Describe how recombinant plasmids are used to transform bacterial cells.

201 202

c

10

Describe the applications of transformed bacterial cells in industry, medicine, and agriculture.

201 202 209 - 211

CL

N AS OT SR F OO OR M US E

8

c


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195 What is DNA Manipulation?

Key Idea: DNA manipulation alters an organism's DNA either by adding new DNA or editing the existing DNA. DNA manipulation (genetic engineering) involves the direct manipulation of an organism's genome using biotechnology. This can be achieved by introducing new DNA into an organism or by editing its existing DNA. DNA manipulation

aims to produce improved or novel organisms with specific desirable traits. Genetic engineering has wide applications in food technology, industry, agriculture, environmental clean up, pharmaceutical production, and vaccine development. Organisms that have had their DNA altered are called genetically modified organisms (or GMOs).

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How are genetically modified organisms produced?

Foreign gene is inserted into host DNA

Existing gene is altered

Host DNA

Host DNA

Gene is deleted or deactivated Host DNA

Add a foreign gene

Alter an existing gene

Delete or ‘turn off’ a gene

A novel (foreign) gene is inserted from another species. This will enable the GMO to express the trait encoded by the new gene. Organisms genetically altered in this way are referred to as transgenic.

An existing gene may be altered to make it express at a higher level (e.g. growth hormone) or in a different way (in tissue that would not normally express it). The technique may provide a way to fix a malfunctioning gene.

An existing gene may be deleted or deactivated (switched off) to prevent the expression of a trait (e.g. the deactivation of the ripening gene in tomatoes produced the Flavr-Savr tomato).

Human insulin, used to treat diabetic patients, is now produced using transgenic bacteria and yeast.

New gene editing technologies, such as CRISPR, are being explored to treat breast cancer (above) and sickle cell disease.

Manipulating gene action is one way in which to control processes such as ripening in fruit.

1. (a) What is DNA manipulation?

2. Describe some of the applications of DNA manipulation:

WEB

KNOW

LINK

LINK

LINK

195 196 197 208

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(b) Using examples, discuss the ways in which an organism may be genetically modified (to produce a GMO):

CL

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196 Making Recombinant DNA

Key Idea: Recombinant DNA (rDNA) is produced by first isolating (or synthesising) a DNA sequence, then inserting it into the genome of a different organism. Recombinant DNA (rDNA) is produced by combining genetic material from two or more different sources. The production of rDNA is possible because the DNA of every organism is

made of the same building blocks (nucleotides). rDNA allows a gene from one organism to be moved into, and expressed in, a different organism. Two important tools are used to create rDNA. Endonucleases (such as restriction enzymes or the CRISPR-Cas9 system) cut the DNA and the enzyme DNA ligase is used to join the sections of DNA together.

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Overview: How is recombinant DNA made?

CDC

Fragments of DNA produced by restriction enzymes (below) are mixed with ethidium bromide, a molecule that fluoresces under UV light. The DNA fragments are then placed on an electrophoresis gel to separate the different lengths of DNA.

Once the DNA fragments are separated, the gel is placed on a UV viewing platform. The area of the gel containing the DNA fragments of the correct length is cut out and placed in a solution that dissolves the gel. This releases the DNA into the solution.

What are restriction enzymes?

``A restriction enzyme is an enzyme that cuts

a double-stranded DNA molecule at a specific recognition site (a specific DNA sequence). There are many different types of restriction enzymes, each has a unique recognition site.

The solution containing the DNA is centrifuged at high speed to separate out the DNA. Centrifugation works by separating molecules of different densities. Once isolated, the DNA can be spliced into another DNA molecule.

Recognition site

Recognition site

G A AT T C

G A AT T C

C T TA A G

C T TA A G

cut

cut

Restriction enzyme cuts here

cut

DNA

``Some restriction enzymes produce DNA fragments

with two sticky ends (right). A sticky end has exposed nucleotide bases at each end. DNA cut in such a way is able to be joined to other DNA with matching sticky ends. Such joins are specific to their recognition sites.

G

A AT T

G

C T TA A

``Some restriction enzymes produce a DNA fragment with two blunt ends (ends with no exposed nucleotide bases). The piece it is removed from is also left with blunt ends. DNA cut in such a way can be joined to any other blunt end fragment. Unlike sticky ends, blunt end joins are non-specific because there are no sticky ends to act as specific recognition sites.

C

G

C T TA

Fragment

G

A

Sticky end

A AT T C

G

DNA fragment with two sticky ends

G

C T TA A

Sticky end

1. What is the purpose of restriction enzymes in making recombinant DNA?

Recognition site

Restriction enzyme cuts here DNA

3. Why is it useful to have many different kinds of restriction enzymes?

Recognition site cut

CCCGGG

CCCGGG

GGGCCC

GGGCCC

cut

cut

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2. Distinguish between sticky end and blunt end fragments:

A AT T C

The cut by this type of restriction enzyme leaves no overhang CCC

LINK GGG CCC

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GGG

CCC

LINK

GGG

WEB GGG CCC

208 204 196

KNOW


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Creating a recombinant DNA plasmid

1

Fragments with matching sticky ends can be joined by basepairing. This process is called annealing. This allows DNA fragments from different sources to be joined.

This other end of the fragment joins to the other sticky end of the plasmid.

TA A C T

Plasmid DNA fragment

A AT T C

G

G

G

CTT AA

Foreign DNA fragment

Hydrogen bonds form between the fragments

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2

Two pieces of DNA are cut by the same restriction enzyme (they will produce fragments with matching sticky ends).

G

The DNA fragments are joined by the enzyme DNA ligase, which catalyses the formation of a phosphodiester bond. This produces a molecule of recombinant DNA.

3

The joined fragments will usually form either a linear or a circular molecule, as shown here (right) as recombinant plasmid DNA.

C

T

T A

A

Detail of restriction site

G

A A T T

C

TA A G C T

Restriction sites on the fragments are attracted by base pairing only

Break in the DNA molecule

Plasmid DNA fragment

G

C

Foreign DNA fragment

A A T T C

T

T A

G

A

DNA ligase

Detail of restriction site

G

A A T T

C

TAA G C T

25kartika

pGLO is a plasmid engineered to contain Green Fluorescent Protein (gfp). pGLO has been used to create fluorescent organisms, including the bacteria above (bright patches on agar plates).

Recombinant plasmid DNA

Fragments linked permanently by DNA ligase

No break in DNA molecule

G

The fragments are joined by the enzyme DNA ligase

C

A A T T C

T

T A

A

G

4. Explain in your own words the two main steps in the process of joining two DNA fragments together:

(a) Annealing:

(b) DNA ligase:

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5. Explain why ligation can be considered the reverse of the restriction digestion process:

CL

6. Why can recombinant DNA be expressed in any kind of organism, even if it contains DNA from another species?

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197 New Tools: Gene Editing with CRISPR for CRISPR to work: an RNA guide that locates and binds to the target piece of DNA and the Cas9 endonuclease that unwinds and cuts the DNA. The technology has potential applications in correcting mutations responsible for disease, switching faulty genes off, adding new genes to an organism, or studying the effect of specific genes. It represents a major advance because it allows more precise and efficient gene editing at much lower cost than ever before. Single guide RNA (sgRNA) is a short synthetic RNA sequence designed to guide Cas9 to the site of interest (e.g. a faulty gene sequence). It contains a nucleotide section which is complementary to the DNA of interest.

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Key Idea: CRISPR is a complex comprising Cas9 endonuclease and sgRNA. The CRISPR complex cuts DNA at very specific sequences and can be used to edit genes. CRISPR-Cas9 (shortened to CRISPR and pronounced crisper) is an endonuclease complex occurring naturally in bacteria, which use it to edit the DNA of invading viruses. CRISPR is able to target specific stretches of DNA and edit it at very precise locations. Two key components are required Cas9 is guided to the target site by sgRNA. Cas9 unwinds the DNA and cuts both strands at a specific point.

Cutting point

Target DNA sequence

5'

3' 5'

5'

3'

The PAM sequence (NGG)* lies directly downstream of the target sequence on the non-target DNA strand. Recognition of PAM by Cas9 destabilises the DNA allowing the sgRNA to be inserted. Cas9 will not function if PAM is absent. *N can be any nucleotide

5'

5'

3' 5'

3'

5'

5'

The cut DNA can be repaired using one of the following methods:

Gene knock in "gene editing"

A new DNA sequence is inserted into the DNA break. For example allows a faulty gene sequence can be replaced with the correct sequence to restore normal gene function.

Gene knock out "gene silencing"

As the cell's normal repair process mend the broken DNA, errors occur resulting in the insertion or deletion of nucleotide bases. The resulting frame-shift mutation changes the way the nucleotide sequence is read, either disabling gene function or producing a STOP signal. This technique can be used to silence a faulty gene.

1. What are the roles of the following in CRISPR gene editing:

(a) Cas9:

(b) sgRNA:

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3. What benefits are offered by CRISPR technology?

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2. Outline two ways CRISPR can be used to edit genes:

LINK

WEB

195 197

KNOW


198 DNA Amplification Using PCR

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288

Key Idea: PCR uses a polymerase enzyme to copy a DNA sample, producing billions of copies in a few hours. Many procedures in DNA technology, e.g. DNA sequencing and profiling, require substantial amounts of DNA yet, very often, only small amounts are obtainable (e.g. DNA from a crime scene or from an extinct organism). PCR (polymerase

chain reaction) is a technique for reproducing large quantities of DNA in the laboratory from an original sample. For this reason, it is often called DNA amplification. The technique is outlined below for a single cycle of replication. Subsequent cycles replicate DNA at an exponential rate, so PCR can produce billions of copies of DNA in only a few hours. DNA polymerase: A thermally stable form of the enzyme is used (e.g. Taq polymerase). . This is extracted from thermophilic bacteria.

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A single cycle of PCR

Direction of synthesis

Primer annealed

Nucleotides

Primer moving into position

A DNA sample (called target DNA) is obtained. It is denatured (DNA strands are separated) by heating at 98oC for 5 minutes.

The sample is cooled to 60oC. Primers are annealed (bonded) to each DNA strand. In PCR, the primers are short strands of DNA; they provide the starting sequence for DNA extension.

Direction of synthesis

Free nucleotides and the enzyme DNA polymerase are added. DNA polymerase binds to the primers and, using the free nucleotides, synthesises complementary strands of DNA.

After one cycle, there are now two copies of the original DNA.

Repeat for about 25 cycles

Repeat cycle of heating and cooling until enough copies of the target DNA have been produced

Loading tray Prepared samples in tiny PCR tubes are placed in the loading tray and the lid is closed.

Temperature control Inside the machine are heating and refrigeration mechanisms to rapidly change the temperature.

Dispensing pipette Pipettes with disposable tips are used to dispense DNA samples into the PCR tubes.

Thermal cycler

Amplification of DNA can be carried out with simple-to-use machines called thermal cyclers. Once a DNA sample has been prepared, in just a few hours the amount of DNA can be increased billions of times. Thermal cyclers are in common use in the biology departments of universities, as well as other kinds of research and analytical laboratories. The one pictured on the left is typical of this modern piece of equipment. DNA quantitation The amount of DNA in a sample can be determined by placing a known volume in this quantitation machine. For many genetic engineering processes, a minimum amount of DNA is required.

WEB

KNOW

LINK

198 206

N AS OT SR F OO OR M US E

1. Explain the purpose of PCR:

CL

RA

Controls The control panel allows a number of different PCR programmes to be stored in the machine’s memory. Carrying out a PCR run usually just involves starting one of the stored programmes.

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2. Describe how the polymerase chain reaction works:

3. Describe three situations where only very small DNA samples may be available for sampling and PCR could be used: (a)

(b)

(c)

4. After only two cycles of replication, four copies of the double-stranded DNA exist. Calculate how much a DNA sample will have increased after:

(a) 10 cycles:

(b) 25 cycles:

5. The risk of contamination in the preparation for PCR is considerable.

(a) Describe the effect of having a single molecule of unwanted DNA in the sample prior to PCR:

(b) Describe two possible sources of DNA contamination in preparing a PCR sample:

Source 1:

Source 2: (c) Describe two precautions that could be taken to reduce the risk of DNA contamination: Precaution 1:

Precaution 2:

(b)

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(a)

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6. Describe two other genetic engineering/genetic manipulation procedures that require PCR amplification of DNA:


199 Gel Electrophoresis

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290

Key Idea: Gel electrophoresis is used to separate DNA fragments on the basis of size. DNA can be loaded onto an electrophoresis gel and separated by size. DNA has an overall negative charge, so when an electrical current is run through a gel, the DNA moves towards the positive electrode. The rate at which the DNA molecules move through the gel depends primarily on their size and the

strength of the electric field. The gel they move through is full of pores (holes). Smaller DNA molecules move through the pores more quickly than larger ones. At the end of the process, the DNA molecules can be stained and visualised as a series of bands. Each band contains DNA molecules of a particular size. The bands furthest from the start of the gel contain the smallest DNA fragments.

Analysing DNA using gel electrophoresis

(-ve)

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(-ve)

DNA solutions: Mixtures of different sizes of DNA fragments are loaded in each well in the gel.

DNA markers, a mixture of DNA molecules with known molecular weights (size) are often run in one lane. They are used to estimate the sizes of the DNA fragments in the sample lanes. The figures below are hypothetical markers (bp = base pairs).

(-ve)

(-ve)

DNA is negatively charged because the phosphates (blue) that form part of the backbone of a DNA molecule have a negative charge.

5 lanes

Negative electrode (–)

Steps in the process of gel electrophoresis of DNA

Wells: Holes are made in the gel with a comb, acting as a reservoir for the DNA solution.

DNA fragments: The gel matrix acts as a sieve for the negatively charged DNA molecules as they move towards the positive terminal. Small fragments move easily through the matrix, whereas large fragments don't.

As DNA molecules migrate through the gel, large fragments will lag behind small fragments. As the process continues, the separation between larger and smaller fragments increases.

Large fragments

50,000 bp 20,000 bp 10,000 bp 5000 bp 2500 bp

Small fragments

1000 bp

500 bp

Tray: The gel is poured into this tray and allowed to set.

Positive electrode (+)

Gel: A gel is prepared, which will act as a support for separation of the fragments of DNA. The gel is a jelly-like material, called agarose.

1. A tray is prepared to hold the gel matrix. 2. A gel comb is used to create holes in the gel. The gel comb is placed in the tray. 3. Agarose gel powder is mixed with a buffer solution (this stabilises the DNA). The solution is heated until dissolved and poured into the tray and allowed to cool. 4. The gel tray is placed in an electrophoresis chamber and the chamber is filled with buffer, covering the gel. This allows the electric current from electrodes at either end of the gel to flow through the gel. 5. DNA samples are mixed with a “loading dye” to make the DNA sample visible. The dye also contains glycerol or sucrose to make the DNA sample heavy so that it will sink to the bottom of the well. 6. The gel is covered, electrodes are attached to a power supply and turned on. 7. When the dye marker has moved through the gel, the current is turned off and the gel is removed from the tray. 8. DNA molecules are made visible by staining the gel with methylene blue or ethidium bromide which binds to DNA and will fluoresce in UV light.

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1. What is the purpose of gel electrophoresis?

(a) (b) 3. Why do the smallest fragments travel through the gel the fastest?

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2. Describe the two forces that control the speed at which fragments pass through the gel:

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200 Interpreting Electrophoresis Gels be used as a profile for a species or individual. Commonly, the gene for cytochrome oxidase I (COXI), a mitochondrial protein, is used to distinguish animal species. The genetic information from this gene is both large enough to measure differences between species and small enough to have the differences make sense (i.e. the differences occur in small regions and aren't hugely varied).

Read in this direction

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Key Idea: The banding pattern on an electrophoresis gel can give information about genetic variation and relationships. Once made, an electrophoresis gel must be interpreted. If a specific DNA base sequence was being investigated, then the band pattern can be used to determine the DNA sequence and the protein that it encoded. Alternatively, depending on how the original DNA was treated, the banding pattern may

T A G T

T

A

G

C

T

Cow

A

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C

Sheep

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Goat

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Horse

1. The photographs above show gel electrophoresis results for four species.

(a) For each of the species determine the sequence of DNA: Cow DNA sequence: Sheep DNA sequence: Goat DNA sequence: Horse DNA sequence: Based on the number of differences in the DNA sequences:

(b) Identify the two species that are most closely related:

(c) Identify the two species that are the least closely related:

2. What makes COXI useful for comparing species by gel electrophoresis?

A

B

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D

E

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Calibration

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3. Determine the relatedness of each individual (A-E) using each banding pattern on the set of DNA profiles (left). When you have done this, complete the phylogenetic tree by adding the letter of each individual.

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201 Using Recombinant Plasmids in Industry

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Key Idea: Inserting useful genes into bacteria to produce biofactories can solve the problem of shortages in the manufacturing and food industries.

The issue ►► Recombinant DNA technology can be used to produce industrially important enzymes in large quantities. ►► Chymosin (also known as rennin) is an enzyme that digests milk proteins. It is the active ingredient in rennet, a substance used by cheesemakers to clot milk into curds. ►► Traditionally rennin is extracted from "chyme", i.e. the stomach secretions of suckling calves (hence its name of chymosin).

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►► By the 1960s, a shortage of chymosin was limiting the volume of cheese produced.

Concept 1

Enzymes are proteins made up of amino acids. The amino acid sequence of chymosin can be determined and the mRNA coding sequence for its translation identified.

►► Enzymes from fungi were used as an alternative but were unsuitable because they caused variations in the cheese flavour.

Concept 2

Concept 3

Concept 4

Reverse transcriptase can be used to synthesise a DNA strand from the mRNA. This process produces DNA without the introns, which cannot be processed by bacteria.

DNA can be cut at specific sites using restriction enzymes and rejoined using DNA ligase. New genes can be inserted into self-replicating bacterial plasmids.

Under certain conditions, bacteria are able to lose or take up plasmids from their environment. Bacteria are readily grown in vat cultures at little expense.

Concept 5

The protein in made by the bacteria in large quantities.

Techniques

The amino acid sequence of chymosin is first determined and the RNA codons for each amino acid identified.

Plasmid isolated from E. coli bacteria.

mRNA matching the identified sequence is isolated from the stomach of young calves. Reverse transcriptase is used to transcribe mRNA into DNA. The DNA sequence can also be made synthetically once the sequence is determined. The DNA is amplified using PCR.

Initially, the gene coding for chymosin was isolated from the stomach of a milk-fed suckling calf (less than 10 days old). Now genes are produced by PCR.

Plasmids from E. coli bacteria are isolated and cut using restriction enzymes. The DNA sequence for chymosin is inserted using DNA ligase.

Plasmid

Plasmids are returned to E. coli by placing the bacteria under conditions that induce them to take up plasmids.

Outcomes

The transformed bacteria are grown in vat culture. Chymosin is produced by E. coli in packets within the cell that are separated during the processing and refining stage.

Transformed bacterial cells are grown in a vat culture

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The recombinant plasmid is taken up by the bacteria.

Recombinant chymosin entered the marketplace in 1990. It established a significant market share because cheesemakers found it to be cost effective, of high quality, and in consistent supply. Most cheese is now produced using recombinant chymosin such as CHY-MAX.

Further applications

A large amount of processing is required to extract chymosin from E.coli. There are now a number of alternative bacteria and fungi that have been engineered to produce the enzyme. Most chymosin is now produced using the fungi Aspergillus niger and Kluyveromyces lactis. Both are produced in a similar way as that described for E. coli.

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Recombinant plasmid

Restriction enzyme cuts the plasmid and DNA ligase joins the chymosin gene into the plasmid DNA.

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Lipase from Aspergillus oryzae is used in processing of palm oil to produce low cost cocoa butter substitutes (above), which have a similar 'mouth feel' to cocoa butter.

Dual Freq

Acetolactate decarboxylase from B. subtilis is an enzyme used in the brewing industry. It reduces maturation time of the beer by by-passing a rate-limiting step.

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Enzymes from GMOs are widely used in the baking industry. Maltogenic alpha amylase from Bacillus subtilis bacteria is used as an anti-staling agent to prolong shelf life. Hemicellulases from B. subtilis and xylanase from the fungus Aspergillus oryzae are used for improvement of dough, crumb structure, and volume during the baking process.

Romain Behar

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1. Describe the main use of chymosin:

2. What was the traditional source of chymosin?

3. Summarise the key concepts that led to the development of the technique for producing chymosin:

(a) Concept 1:

(b) Concept 2:

(c) Concept 3:

(d) Concept 4:

(e) Concept 5:

4. Discuss how the gene for chymosin was isolated and how the technique could be applied to isolating other genes:

5. Describe three advantages of using chymosin produced by GE bacteria over chymosin from traditional sources: (a)

(c)

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(b)

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6. Explain why the fungus Aspergillus niger is now more commonly used to produce chymosin instead of E. coli:


202 Using Recombinant Plasmids in Medicine

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The Issue ►► Type I diabetes mellitus is a metabolic disease

Key Idea: By using microorganisms to make human insulin, problematic issues of cost, allergic reactions, and ethics have been addressed.

caused by a lack of insulin. Around 25 people in every 100,000 suffer from type I diabetes.

►► It is treatable only with injections of insulin. ►► In the past, insulin was taken from the

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pancreases of cows and pigs and purified for human use. The method was expensive and some patients had severe allergic reactions to the foreign insulin or its contaminants.

Insulin A chain

Insulin B chain

Concept 1

Concept 2

Concept 3

DNA can be cut at specific sites using restriction enzymes and joined together using DNA ligase. Genes can be inserted into self-replicating bacterial plasmids at the point where the cuts are made.

Plasmids are small, circular pieces of DNA found in some bacteria. They usually carry genes useful to the bacterium. E. coli plasmids can carry promoters required for the transcription of genes.

Under certain conditions, Bacteria are able to lose or pick up plasmids from their environment. Bacteria can be readily grown in vat cultures at little expense.

Concept 4

The DNA sequences coding for the production of the two polypeptide chains (A and B) that form human insulin can be isolated from the human genome.

Techniques

The gene is chemically synthesised as two nucleotide sequences, one for the insulin A chain and one for the insulin B chain. The two sequences are small enough to be inserted into a plasmid.

The nucleotide sequences for each insulin chain are synthesised separately and placed into separate plasmids

Plasmids are extracted from Escherichia coli. The gene for the bacterial enzyme b-galactosidase is located on the plasmid. To make the bacteria produce insulin, the insulin gene must be linked to the b-galactosidase gene, which carries a promoter for transcription. Restriction enzymes are used to cut plasmids at the appropriate site and the A and B insulin sequences are inserted. The sequences are joined with the plasmid DNA using DNA ligase.

The recombinant plasmids are introduced into the bacterial cells

b-galactosidase + chain A

The gene is expressed as separate chains

b-galactosidase + chain B

The recombinant plasmids are inserted back into the bacteria by placing them together in a culture that favours plasmid uptake by bacteria. The bacteria are then grown and multiplied in vats under carefully controlled growth conditions.

Outcomes

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Insulin A chain

The product consists partly of b-galactosidase, joined with either the A or B chain of insulin. The chains are extracted, purified, and mixed together. The A and B insulin chains connect via disulfide cross linkages to form the functional insulin protein. The insulin can then be made ready for injection in various formulations.

Further Applications

Insulin B chain

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Disulfide bond

The techniques used to produce human insulin from genetically modified bacteria can be applied to a range of human proteins and hormones. Proteins currently being produced include human growth hormone, interferon, and factor VIII.

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Insulin production in Saccharomyces

The gene for human insulin is inserted into a plasmid. The yeast plasmid is larger than that of E.coli, so the entire gene can be inserted in one piece rather than as two separate pieces.

Cleavage site

The proinsulin protein that is produced folds into a specific shape and is cleaved by the yeast's own cellular enzymes, producing the completed insulin chain.

S

S

S

S

S

By producing insulin this way, the secondary step of combining the separate protein chains is eliminated, making the refining process much simpler.

S

S

S

S

S

S

S

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Yeast cells are eukaryotic and hence are much larger than bacterial cells. This enables them to accommodate much larger plasmids and proteins within them.

Cleavage site

1. Describe the three major problems associated with the traditional method of obtaining insulin to treat diabetes: (a) (b) (c)

2. Explain the reasoning behind using E. coli to produce insulin and the benefits that GM technology has brought to diabetics:

3. Explain why, when using E. coli, the insulin gene is synthesised as two separate A and B chain nucleotide sequences:

4. Why are the synthetic nucleotide sequences (‘genes’) 'tied' to the b-galactosidase gene?

(b) Secretion and purification of the protein product:

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5. Yeast (Saccharomyces cerevisiae) is also used in the production of human insulin. Discuss the differences in the production of insulin using yeast and E. coli with respect to: (a) Insertion of the gene into the plasmid:


203 KEY TERMS AND IDEAS: Did You Get It?

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1. Match each term to its definition, as identified by its preceding letter code. annealing

A An organism that has had part of its DNA sequence altered either by the removal

blunt end

or insertion of a piece of DNA.

B The site or sequence of DNA at which a restriction enzyme attaches and cuts.

CRISPR-Cas9 system

C An organism or artificial vehicle that is capable of transferring a DNA sequence to another organism.

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DNA amplification

D The repairing or attaching of fragmented DNA by ligase enzymes.

E A type of cut in a length of DNA caused by a restriction enzyme that results in

DNA ligation

DNA polymerase

both strands of DNA being the same length.

F A process that is used to separate different lengths of DNA by placing them in a

gel matrix placed in a buffered solution through which an electric current is passed.

gel electrophoresis

G An enzyme that is able to cut a length of DNA at a specific sequence or site.

genetic modification

H A genome editing tool used to edit DNA at a specific sequence.

I The pairing (by hydrogen bonding) of complementary single-stranded nucleic

GMO

acids to form a double-stranded polynucleotide. The term is applied to making recombinant DNA, to the binding of a DNA probe, or to the binding of a primer to a DNA strand during PCR.

plasmid

J A cut in a length of DNA by a restriction enzyme that results in two strands of DNA being different lengths with one strand overhanging the other.

polymerase chain reaction

K A small circular piece of DNA commonly found in bacteria.

primer

L The process of producing more copies of a length of DNA, normally using PCR.

recognition site

M The alteration of an organism's DNA either by adding new DNA or editing the existing DNA.

recombinant DNA

N A reaction that is used to amplify fragments of DNA using cycles of heating and cooling.

recombinant plasmid

O DNA that has had a new sequence added so that the original sequence has been changed.

restriction enzyme

P An enzyme that is able to replicate DNA and commonly used in PCR to amplify a length of DNA.

sticky end

Q A thermally stable form of polymerase isolated from thermophilic bacteria.

taq polymerase

R A short length of DNA used to identify the starting sequence for PCR so that polymerase enzymes can begin amplification.

vector

S A plasmid that has foreign DNA inserted into it.

2. Below is a DNA sequence of sections, A, B, C, D, and E and A', B', C', D', and E'. A scientist wants to isolate sections B, C, and D as a continuous group by PCR. Primers are B and D'. Complete the PCR process, using the right hand (') strand to show how B, C, and D are isolated. The first step is done for you. Note that the process could be shown for the left hand strand also but the sequence of events would be different.

A

A'

Primer

A'

B

B'

B

B'

B

B'

C

C'

C'

C

C'

D

D'

D'

D

D'

E

E'

E'

E

E'

TEST

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Isolation using LH strand

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Unzipped

A'

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Biological knowledge and society

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Unit 4 Outcome 2

Key terms

Genetic techniques

DNA (genetic) profiling

Key knowledge

forensics

1

c

i

IN-VIVO GENE CLONING: Describe and explain gene cloning using plasmids. Describe one or more of the applications of molecular clones.

c

ii

PROFILING: Describe and explain DNA profiling using PCR. Discuss the applications of DNA profiling as a forensic and diagnostic tool.

c

iii

GENETIC SCREENING: Describe and explain the use of genetic screening in the diagnosis and future management of heritable disorders.

GMO

marker gene

microsatellite

molecular clone

Outline the techniques i-iii below that apply knowledge of DNA manipulation. Include the social and ethical issues involved with each.

c

gene technology

genetic modification

Activity number

204

205 206 207

plasmid

recombinant DNA technology

reverse transcription STR

transgenic organism transgenesis vector

NIH

Genetically modified and transgenic organisms Key knowledge

c

2

Distinguish between genetically modified and transgenic organisms, recognising one as a subset of the other. Use your knowledge of DNA to explain why it is possible to create transgenic organisms.

c

3

Describe the use of GMOs and transgenic organisms in agriculture to change crop characteristics, increase crop production or yield, or confer resistance to pests or diseases.

c

4

Discuss the biological, social, and ethical implications that arise as a result of GMO use. Discussion could include (but is not restricted to) effect on developing nations, food labelling and safety issues, GM contamination and spread of transgenes, and consumer choice.

Responding to the threat of new diseases Key knowledge

CDC

Activity number 208

209 - 211

212

Activity number 213

5

Distinguish between epidemics and pandemics.

c

6

Describe strategies for responding to the emergence of new diseases in today's world, where international travel and trade connects populations globally.

213 214 215

c

7

Explain how pathogens are identified and diseases are treated based on sound scientific knowledge and principles.

216

Drugs against pathogens Key knowledge

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c

Activity number

8

Explain the concept of rational drug design with reference to the anti-influenza drug Relenza. Include reference to the complementary nature of drug molecules that bind and inhibit enzyme activity.

c

9

Describe how chemical agents (drugs) are used against pathogens. Distinguish between antibiotics and antiviral drugs with reference to their mode of action and efficacy against the pathogen.

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217 218

219


204 In Vivo Gene Cloning

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host organism, they must have one or more sites at which a restriction enzyme can cut, and they must have some kind of genetic marker that allows them to be identified. Bacterial plasmids are commonly used vectors because they are easy to manipulate, their restriction sites are well known, and they are readily taken up by cells in culture. Once the molecular clone has been taken up by bacterial cells, and those cells are identified, the gene can be replicated (cloned) many times as the bacteria grow and divide in culture.

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Key Idea: In vivo cloning describes the insertion of a gene into an organism and using the replication machinery of that organism to multiply the gene or produce its protein product. Recombinant DNA techniques (restriction digestion and ligation) are used to insert a gene of interest into the DNA of a vector (e.g. plasmid or viral DNA). This produces a recombinant DNA molecule called a molecular clone that can transmit the gene of interest to another organism. To be useful, all vectors must be able to replicate inside their

Cloning a human gene

Bacterial cell. e.g. E. coli

Human cell

Plasmid

Chromosome

DNA in chromosome A gene of interest (DNA fragment) is isolated and prepared by removal of introns.

Human gene

Sticky end

Restriction enzyme recognition sequence

Tetracyclineresistance gene

Sticky end

Plasmid vector

Promoter and terminator sequences added

The recombinant plasmid, or molecular clone, is introduced into a bacterial cell by adding the DNA to a bacterial culture. Under the right conditions, some bacteria will take up the plasmid from solution by the process of transformation.

Colonies of bacteria that carry the recombinant plasmid can be identified by the fact that they are resistant to ampicillin but sensitive to tetracycline. These colonies can be isolated and grown in culture.

Agar plate with bacterial colonies. Only some have the plasmid with the human gene.

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Human gene

Recombinant DNA molecule

A commercially available plasmid engineered to contain certain restriction sites is used in the cloning process.

The restriction enzyme cuts the plasmid DNA at its single recognition sequence, disrupting, for example, the tetracycline resistance gene.

The DNA fragments are mixed together and the complementary sticky ends are attracted by basepairing. The enzyme DNA ligase is added to bond the sticky ends.

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Sticky ends

Insulin is now commonly made using recombinant DNA. The gene for insulin is placed into a bacterial plasmid. As the bacteria replicates it also replicates the plasmid. Huge vats of insulin producing bacteria can quickly be grown. Other human proteins, e.g. factor VII and adenosine deaminase, are produced the same way.

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Both the human DNA and the plasmid are treated with the same restriction enzyme to produce identical sticky ends.

Ampicillinresistance gene

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gfp as a gene marker

Antibiotic resistant marker genes may be used to identify the bacteria that have taken up the foreign (e.g. human) DNA. The plasmid used often carries two genes that provide the bacteria with resistance to the antibiotics ampicillin and tetracycline. Without this plasmid, the bacteria have no antibiotic resistance genes. A single restriction enzyme recognition sequence lies within the tetracycline resistance gene. A foreign gene, spliced into this position, will disrupt the tetracycline resistance gene, leaving the bacteria vulnerable to this antibiotic. It is possible to identify the bacteria that successfully take up the recombinant plasmid by growing the bacteria on media containing ampicillin, and transferring colonies to media with both antibiotics.

Most often today, another gene acts as a marker instead of the tetracycline resistance gene. The gene for Green Fluorescent Protein (gfp below), isolated from the jellyfish Aequorea victoria, has become well established as a marker for gene expression in the recombinant organism. The gfp gene is recombined with the gene of interest and transformed cells can then be detected by the presence of the fluorescent product (cells with gfp present glow green under fluorescent light).

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Antibiotic resistance as a marker

1. Explain why it might be desirable to use in vivo methods to clone genes rather than PCR:

2. Explain when it may not be desirable to use bacteria to clone genes:

3. Explain how a human gene is removed from a chromosome and placed into a plasmid:

4. A bacterial plasmid replicates at the same rate as the bacteria. If a bacteria containing a recombinant plasmid replicates and divides once every thirty minutes, calculate the number of plasmid copies there will be after twenty four hours:

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5. When cloning a gene using plasmid vectors, the bacterial colonies containing the recombinant plasmids are mixed up with colonies that have none. All the colonies look identical, but some have taken up the plasmids with the human gene, and some have not. Explain how the colonies with the recombinant plasmids are identified:

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6. Explain why the gfp marker is a more desirable gene marker than genes for antibiotic resistance:


205 DNA Profiling Using PCR

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DNA profiling, which identifies the natural variations found in every person’s DNA. Identifying these DNA differences is a useful tool for forensic investigations. In Victoria, the Victoria Police Forensic Services Department (FSD) is one of the largest providers of forensic science services in Australia. The FSS targets 9 STR sites; enough to guarantee that the odds of someone else sharing the same result are extremely unlikely (about one in a billion). DNA profiling has been used to help solve current and previously unsolved crimes and can also be used to establish genetic relatedness (e.g. in paternity disputes), or when testing for disease.

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Key Idea: Short units of DNA that repeat a variable number of times in different people can be used to produce individual genetic profiles. In chromosomes, some of the DNA contains simple, repetitive sequences. These non-coding sequences are found scattered throughout the genome. Some repeating sequences, called microsatellites or short tandem repeats (STRs), are very short (2-6 base pairs) and can repeat up to 100 times. The human genome has many different STRs. Equivalent sequences in different people vary considerably in the numbers of the repeating unit. This phenomenon is used

Telomeres

Microsatellites (Short (short Tandem tandem Repeats) repeats) Microsatellites

Microsatellites consist of a variable number of tandem repeats of a 2 to 6 base pair sequence. In the example below it is a two base sequence (CA) that is repeated.

Centromeres

Homologous pair of chromosomes

The human genome contains about 100 000 separate blocks of tandem repeats of the dinucleotide: CA. One such block at a known location on a chromosome is shown below: DNA

CA CA CA CA CA CA CA CA CA CA CA

DNA

Microsatellites are found throughout the genome: within genes (introns) and between genes, and particularly near centromeres and telomeres.

Flanking regions to which PCR primers can be attached

How short tandem repeats are used in DNA profiling

DNA from individual ‘A’:

The tandem repeat may exist in two versions (alleles) in an individual; one on each homologous chromosome. Each of the strands shown left is a double stranded DNA, but only the CA repeat is illustrated.

DNA from individual ‘B’:

This diagram shows how three people can have quite different microsatellite arrangements at the same point (locus) in their DNA. Each will produce a different DNA profile using gel electrophoresis:

DNA from individual ‘C’:

Microsatellite

Extract DNA from sample

A sample collected from the tissue of a living or dead organism is treated with chemicals and enzymes to extract the DNA, which is separated and purified.

Microsatellite from individual ‘A’:

Primers

Flanking region

Microsatellite from individual ‘B’:

Microsatellite from individual ‘C’:

STR

DNA

Amplify microsatellite using PCR

The results of PCR are many fragments

A

B

C

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Specific primers (arrowed) that attach to the flanking regions (light grey) either side of the microsatellite are used to make large quantities of the micro-satellite and flanking regions sequence only (no other part of the DNA is amplified/replicated).

Largest fragments

The fragments are separated by length, using gel electrophoresis. DNA, which is negatively charged, moves toward the positive terminal. The smaller fragments travel faster than larger ones.

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The products of PCR amplification (making many copies) are fragments of different sizes that can be directly visualised using gel electrophoresis.

Smallest fragments

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Visualise fragments on a gel

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The photo above shows a film output from a DNA profiling procedure. Those lanes with many regular bands are used for calibration; they contain DNA fragment sizes of known length. These calibration lanes can be used to determine the length of fragments in the unknown samples.

DNA profiling can be automated in the same way as DNA sequencing. Computer software is able to display the results of many samples run at the same time. In the photo above, the sample in lane 4 has been selected and displays fragments of different length on the left of the screen.

1. Describe the properties of short tandem repeats that are important to the application of DNA profiling technology:

2. Explain the role of each of the following techniques in the process of DNA profiling:

(a) Gel electrophoresis:

(b) PCR:

3. Describe the three main steps in DNA profiling using PCR: (a)

(c)

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(b)

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4. Explain why as many as 10 STR sites are used to gain a DNA profile for forensic evidence:


206 Forensic Applications of DNA Profiling

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DNA evidence has been used to identify body parts, solve cases of industrial sabotage and contamination, for paternity testing, and even in identifying animal products illegally made from endangered species. DNA is isolated and profiles are made from all samples and compared to known DNA profiles such as that of the victim.

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Key Idea: DNA profiling has many forensic applications, from identifying criminal offenders to saving endangered species. The use of DNA as a tool for solving crimes such as homicide is well known, but it can also has several other applications.

Offender was wearing a cap but lost it when disturbed. DNA can be retrieved from flakes of skin and hair.

DNA left behind when offender drunk from a cup in the kitchen.

A

B

Calibration

C

D

Profiles of collected DNA

Victim (D)

Investigator (C)

Unknown DNA samples are compared to DNA databases of convicted offenders and to the DNA of the alleged offender.

Bloodstain. DNA can be extracted from white blood cells in the sample

Hair. DNA can be recovered from cells at the base of the strand of hair.

The role of frequency and probability

Every person has two copies of each chromosome and therefore two copies (alleles) of every testable DNA marker. For example, the short tandem repeat (STR) known as CSF1PO contains between 7 and 15 repeats of GATA and has 9 possible alleles. Some alleles (and therefore genotypes) are more common in the population that others. For the CSF1PO STR, the frequency of the genotype 10,11 (allele 10 and allele 11) is 0.1270, i.e. it appears in 12.7% of the population. When DNA is tested, a number of STRs are sampled (the exact number varies between countries). When the data from all STRs is considered, levels of probability that the DNA came from a certain person can be calculated to 1 in 500 trillion.

A

Alleged offender

During the initial investigation, samples of material that may contain DNA are taken for analysis. At a crime scene, this may include blood and body fluids as well as samples of clothing or objects that the offender might have touched. Samples from the victim are also taken to eliminate them as a possible source of contamination.

Calibration

E

F

G

Profiles from DNA database

Although it does not make a complete case, DNA profiling, in conjunction with other evidence, is one of the most powerful tools in identifying offenders or unknown tissues.

Allele frequencies of the CSF1PO STR Allele (number of repeats)

Frequency

Allele (number of repeats)

Frequency

7

0.0232

12

0.3446

8

0.0212

13

0.0656

9

0.0294

14

0.0092

10

0.2321

15

0.0010

11

0.2736

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1. Why are DNA profiles obtained for both the victim and investigator?

3. What is the frequency of the following CSF1PO alleles: (a) 9: (b) 12: WEB

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2. Use the evidence to decide if the alleged offender is innocent or guilty and explain your decision:

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Paternity testing

Whale DNA: tracking illegal slaughter

DNA profiling can be used to determine paternity (and maternity) by looking for matches in alleles between parents and children. This can be used in cases such as child support or inheritance. DNA profiling can establish the certainty of paternity (and maternity) to a 99.99% probability of parentage. Every STR allele is given the number of its repeats as its name, e.g. 8 or 9. In a paternity case, the mother may be 11, 12 and the father may be 8, 13 for a particular STR. The child will have a combination of these. The table below illustrates this: Mother's alleles

Child's alleles

Father's alleles

CSF1PO

7, 8

8, 9

9, 12

D10S1248

14, 15

11, 14

10, 11

D12S391

16, 17

17, 17

17, 18

D13S317

10, 11

9, 10

8, 9

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The frequency of the each allele occurring in the population is important when determining paternity (or maternity). For example, DNA marker CSF1PO allele 9 has a frequency of 0.0294 making the match between father and child very significant (whereas allele 12 has a frequency of 0.3446, making a match less significant). For each allele, a paternity index (PI) is calculated. These indicate the significance of the match. The PIs are combined to produce a probability of parentage. 10-13 different STRs are used to identify paternity. Mismatches of two STRs between the male and child is enough to exclude the male as the biological father.

Under International Whaling Commission regulations, some species of whales can be captured for scientific research and their meat sold legally. Most, including humpback and blue whales, are fully protected and to capture or kill them for any purpose is illegal. Between 1999 and 2003 Scott Baker and associates from Oregon State University's Marine Mammal Institute investigated whale meat sold in markets in Japan and South Korea. Using DNA profiling techniques, they found around 10% of the samples tested were from fully protected whales including western grey whales and humpbacks. They also found that many more whales were being killed than were being officially reported.

4. For the STR D10S1248 in the example above, what possible allele combinations could the child have?

5. A paternity test was carried out and the abbreviated results are shown below:

DNA marker

Mother's alleles

Child's alleles

Man's alleles

CSF1PO

7, 8

8, 9

9, 12

D10S1248

14, 15

11, 14

10, 11

D19S433

9, 10

10,15

14, 16

D13S317

10, 11

9, 10

8, 9

D2S441

7, 15

7, 9

14, 17

(a) Could the man be the biological father?

(b) Explain your answer:

(b) How could DNA profiling be used to refute official claims of the number of whales captured and sold in fish markets?

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6. (a) How could DNA profiling be used to refute official claims of the type of whales captured and sold in fish markets?


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207 Genetic Screening and Testing

Key Idea: Gene screening uses multiple methods for identifying genetically linked diseases in people. Genetic testing provides specific DNA analysis of a targeted gene. Genetic screening is not a consistent term and is often used interchangeably with genetic testing. However genetic screening is a systematic, wide ranging programme, usually offered by the government, to test asymptomatic individuals

for specific genetically linked disorders (e.g. phenylketonuria or PKU). The methods are varied but usually simple, and give a simple positive or negative result. Genetic testing is the use of DNA analysis by probes, PCR, and other techniques to analyse a DNA sample and provide specific information it, e.g. the type of mutation carried by an individual. This may be carried out before birth (prenatal genetic testing).

Genetic screening programmes

Genetic screening continuum

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These are usually population wide programmes and are used as early diagnosis test, often soon after birth. They include the following tests:

ff Tests for current conditions with a known genetic cause. ff Predictive screening, i.e. screening for genes that may

Population screening

Clinical practice

cause diseases in an individual's future.

ff Carrier screening: testing to find if an asymptomatic person carries a recessive disorder.

Some of these programs are mandatory (e.g. PKU, hypothyroidism, and cystic fibrosis) and are performed within 48 hours of birth. Others are voluntary programmes, for example, for people who may carry genetic disorder and want to establish the risks of passing it to their offspring. Sometimes the genetic screening programmes are undertaken to provide data about the frequency of a genetic disorder in the wider population (e.g. cystic fibrosis has a carrier frequency of about 1 in 25 people).

Mass screening

Opportunistic screening

Cascade screening

Genetic testing

Genetic screening tests can be placed on a continuum. The wider ranging less detailed tests are used for mass screening of the population. More specific tests are done in specialist clinics and when needed. Opportunistic testing is performed when individuals enter the health system for an unrelated illness and tests show possible other problems. Cascade screening is the systematic testing of individuals related to a person identified as a carrier or sufferer of a genetic disorder.

An example of a genetic screening programme

PKU is tested for using blood obtained from pricking a newborn's heel (heel prick test) and collected on filter paper. The most common test for PKU is the Guthrie test. A small disk of the dried blood is placed on an agar plate containing a substance that inhibits bacteria growth. If the blood contains elevated levels of phenylalanine, the bacteria begin to grow, allowing PKU to be detected.

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Phenylketonuria is a metabolic disorder with serious health and mental consequences if it is not treated. People with PKU can not metabolise phenylalanine, a common amino acid. The disorder is not curable but can be managed by controlling phenylalanine intake in the diet. There may be no symptoms in people who are diagnosed early enough.

The heel prick collects blood for PKU and other genetic disorders of amino acid metabolism, congenital hypothyroidism, cystic fibrosis, galactosaemia, and fatty acid oxidation defects.

1. Contrast genetic screening and genetic testing:

2. Why is a simple test used for screening, while a more detailed test is used for genetic testing?

(a) Opportunistic screening:

(b) Cascade screening:

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3. Explain why each of the following would be used:

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Genetic testing: prenatal diagnosis of PKU An example of the use of genetic screening, cascade screening, and genetic testing can be illustrated by the case study of a couple in Bangalore in 2005 who already had one child with PKU and wanted testing of the current pregnancy. Both the mother and father were carriers and therefore heterozygous for the condition. PKU is an autosomal recessive disorder, so two affected alleles are required for the disorder to occur. 2. The DNA corresponding to the phenylalanine hydroxylase gene was amplified using PCR. Three tests were performed on the DNA using short tandem repeats (STR), restriction fragment length polymorphisms (RFLP), and variable number tandem repeats (VNTR).

3. STRs are short sections of repeated DNA, e.g. GATAGATAGATA. The repeat may occur 2-6 times. RFLPs are lengths of DNA cut using restriction enzymes. VNTRs are similar to STRs but are generally longer. The variants of these repetitive sequences act as inherited 'alleles', allowing them to be used as markers for identification. The likelihood of two unrelated individuals having the same allelic pattern is extremely improbable.

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1. DNA was extracted from blood cells from the father, the mother, and the affected child, and cells were taken from the amniotic fluid around the fetus.

GATA GATA GATA GATA A short tandem repeat

DNA

Primer

4. The tests were run of electrophoresis gel. The results for the VNTR and STR gels are shown below. The 'alleles' are not true alleles because they are sequences within the gene: Fa

AC

F

Fa

AC

F

L

L

Allele a

Allele b

Allele 1

Adapted from Sudha Kohli et al

M

M

Allele 2

Two VNTR alleles were produced. The diagram above shows the alleles present as seen on the electrophoresis gel. M = mother, Fa = father, AC = affected child, F = fetus, and L is the test ladder. The affected child and father show one bar, showing they are homozygous. The mother and fetus have two bars and are heterozygous (carriers).

Two STR alleles were also produced. The diagram above shows the alleles present as seen on the electrophoresis gel. The STR marker shows two alleles. Both the parents were heterozygotes. The affected child was homozygous. The fetus was heterozygous. From these two tests alone we can see that the fetus will be a carrier and unaffected. The genetic diagnosis was confirmed using the Guthrie test when the child was born.

4. Why is the DNA of the affected child tested in the prenatal PKU test above?

5. (a) Which alleles did the affected child inherit from the mother?

(b) Which alleles did the affected child inherit from the father?

(b) Explain your reasoning:

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6. (a) Using the electrophoresis profiles above, which parent is the fetus likely to have received the recessive allele from?


208 What is Transgenesis?

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Key Idea: Transgenesis is the insertion of a gene from one species into another, so its protein product is expressed in the second species. Transgenesis has many applications in agriculture, and food and medical technologies. Transgenesis refers to the specific genetic engineering technique of inserting a gene from one species into another

that does not normally contain the gene. It allows direct modification of a genome so that novel traits can be introduced to an organism. Organisms that have undergone transgenesis are called transgenic organisms. The genes are inserted using vectors or by direct insertion of the DNA. The most commonly used vectors are shown below:

Examples of vectors for gene transfer Plasmids

Liposomes

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Viruses

Retrovirus

Recombinant plasmids contain DNA from one or more other organisms

Adenovirus

Lipid bilayer

Novel gene

Viruses are well known for their ability to insert DNA into a host cell. For this reason they have become a favoured tool in transgenesis. Different types of viruses integrate their DNA into the host in different ways. This allows scientists to control where and for how long the new DNA is expressed in the host. However, the size of the piece of DNA that can be transferred is limited to about 8 kb. Also, integration of the DNA into the host DNA can cause unexpected side effects depending on where in the host's chromosome the DNA inserts itself.

Plasmids are circular lengths of DNA that can be up to 1000 kb long (1 kb = 1000 bp). Recombinant plasmids are frequently used to produce transgenic organisms, especially bacteria. The bacteria maybe the final target for the recombinant DNA (e.g. transgenic E. coli producing insulin) or it can be used as a vector to transfer the DNA to a different host (e.g. Agrobacterium tumefaciens is used to transfer the Ti plasmid to plants).

Liposomes are spherical bodies of lipid bilayer. They can be quite large and targeted to specific types of cell by placing specific receptors on their surfaces. Because of their size, liposomes can carry plasmids 20 kb or more, but are less efficient than viruses at transferring the plasmid into a target cell. Liposomes are sometimes used to transfer new DNA into sperm cells, which will then transfer their DNA to the zygote. This is a useful method in birds, where the embryo is enclosed in a shell. However it is not particularly efficient.

Creating a transgenic animal

Creating transgenic mice using pronuclear injection

Pronuclear injection

A gene that has been transferred into another organism is called a transgene. Genes can be introduced directly into an animal cell by microinjection. Multiple copies of the desired transgene are injected via a glass micropipette into a recently fertilised egg cell, which is then transferred to a surrogate mother. Transgenic mice and livestock are produced in this way. However, the process is inefficient: only 2-3% of eggs give rise to transgenic animals and only a proportion of these animals express the transgene adequately.

2b Micropipette injects rat growth hormone gene into a fertilised egg.

This example outlines a successful experiment that used DNA microinjection technology to produce the world’s first transgenic animal.

Transformed egg is cultured

3b to an embryo, then implanted in a surrogate mother.

Mouse B Weight 44 g

Micropipette injects gene

Two eggs are removed from a single female mouse and are fertilised artificially in a test tube.

Mouse A Weight 29 g

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1

Egg cell

4

2a This fertilised egg is unaltered.

Egg nucleus Blunt holding pipette

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Normal egg is cultured to

3a an embryo, then implanted in a surrogate mother.

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The two mice above are siblings, but mouse B is a transgenic organism. A rat growth hormone gene was introduced into its genome.

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Transduction is the transfer of DNA into a bacterium by a virus. Bacteriophages (viruses that infect bacteria) are commonly used to integrate recombinant DNA into a target bacterium.

Zephyris cc 3.0

BioRad/RA

Transfection is the deliberate, often non-viral, introduction of foreign DNA into a cell. There are numerous methods including electroporation and the use of the gene gun (above).

Electroporation cuvettes

Electroporation is a method in which an electric field is applied to cells, causing the plasma membrane to become more permeable. This allows DNA to cross the plasma membrane.

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Transformation is the direct uptake of foreign DNA and is common in bacteria. Recombinant DNA plasmids are mixed with bacteria and the bacteria that take up the DNA are used.

Dr Graham Beards, cc 3.0

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Transferring the DNA

1. What is transgenesis?

2. (a) Describe a feature of viruses that make them well suited as vectors for DNA transfer:

(b) Identify two problems with using viral vectors for DNA transfer:

3. Describe how plasmids are used to transfer new genes to plants:

4. Describe an advantage of liposomes over plasmids and viruses in transgenesis:

5. (a) Describe pronuclear injection:

(b) Describe how pronuclear injection can be used to produce a transgenic mouse:

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6. Describe the difference between, transformation, transduction, and transfection:

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209 The Applications of Transgenesis

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Key Idea: GMOs are used in the food industry, agriculture, horticulture, medicine, and environmental practices. Techniques for genetic manipulation are now widely applied throughout modern biotechnology: in food and enzyme technology, in industry and medicine, and in agriculture and horticulture. Microorganisms are among the most widely

used GMOs, with applications ranging from pharmaceutical production and vaccine development to environmental clean-up and rehabilitation. Crop plants are also commonly genetically modification because they are easily propagated and the potential gains are great. However their use is controversial because transgenes are easily spread between plant species.

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1. Suggest one economic advantage of extending shelf life in fresh produce:

Extending shelf life: Shelf life in fresh produce (e.g. tomatoes) can be extended by switching off the genes for specific enzymes involved in the fruit ripening process (e.g. the enzymes involved in softening of the fruit wall or controlling the production of ethylene).

Pest or herbicide resistance: Plants

2. Suggest one disadvantage of engineering crop plants to be herbicide resistant:

can be engineered to carry and express genes for insect toxins or herbicide resistance. Pest resistant crops do not require spraying and herbicide resistance allows the grower to control weeds without damaging the crops.

3. What is the advantage of using GE bacteria to produce a human hormone like insulin?

Biofactories: Transgenic bacteria

are widely used to produce desirable commodities, such as hormones or proteins. Large quantities of a product can be produced using bioreactors. One example includes injectable human insulin produced by recombinant bacteria or yeast (above).

Vaccine development: Vaccines can be made using gene technology. Genes encoding antigenic vaccine components (e.g. viral proteins) are inserted into a bacterial cell, which then expresses the genes. The gene product is purified and generates an immune response without the risk of ever causing the disease.

4. Why might it be safer to produce a vaccine using gene technology, rather than the pathogen itself?

5. What advantages could be gained by developing a GE crop that produces more protein?

can be engineered to thrive on waste products, such as liquefied newspaper pulp or oil. They degrade pollutants and wastes, and minerals (e.g. mercury) may also be recovered from the bacteria after the cleanup.

Livestock improvement using transgenic animals: Transgenic sheep have been used to enhance wool production in flocks (above, left). The keratin protein of wool is largely made of a single amino acid, cysteine. Injecting developing sheep with the genes for the enzymes that generate cysteine produces woollier sheep. Transgenic sheep carrying the human gene for a protein, Îą-1-antitrypsin, produce the protein in their milk. The antitrypsin is extracted from the milk and can be used to treat hereditary emphysema.

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6. Using animals as live biofactories to produce a valuable product (e.g. a human protein) is controversial. What welfare issues could be associated with such practices?

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Environmental clean-up: Bacteria

technology is now an integral part of the development of new crop varieties. Crops can be engineered to produce higher protein or vitamin levels (e.g. golden rice) or to grow in inhospitable conditions (e.g. salty or arid land).

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Crop improvement: Gene

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210 Engineering for Improved Nutrition The issue  Beta-carotene (β-carotene) is a precursor to vitamin A which is involved in many functions including vision, immunity, fetal development, and skin health.  Vitamin A deficiency is common in developing countries where up to 500,000 children suffer from night blindness, and death rates due to infections are high due to a lowered immune response.  Providing enough food containing useful quantities of β-carotene is difficult and expensive in many countries.

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Key Idea: The use of recombinant DNA to build a new metabolic pathway has greatly increased the nutritional value of a variety of rice.

Concept 1

Rice is a staple food in many developing countries. It is grown in large quantities and is available to most of the population, but it lacks many of the essential nutrients required by the human body for healthy development. It is low in b-carotene.

Concept 2

Concept 3

Concept 4

Rice plants produce b-carotene but not in the edible rice endosperm. Engineering a new biosynthetic pathway would allow b-carotene to be produced in the endosperm. Genes expressing enzymes for carotene synthesis can be inserted into the rice genome.

The enzyme carotene desaturase (CRT1) in the soil bacterium Erwinia uredovora, catalyses multiple steps in carotenoid biosynthesis. Phytoene synthase (PSY) overexpresses a colourless carotene in the daffodil plant Narcissus pseudonarcissus.

DNA can be inserted into an organism's genome using a suitable vector. Agrobacterium tumefaciens is a gall-forming bacterial plant pathogen that is commonly used to insert novel DNA into plants.

The development of golden rice

Techniques

The PSY gene from daffodils and the CRT1 gene from Erwinia uredovora are sequenced.

Ti plasmid

PSY is needed for the synthesis of a colourless carotene.

PSY

T1 CR

The tumour-inducing Ti plasmid is modified to delete the gallforming gene and insert the genes of interest. The parts of the Ti plasmid required for plant transformation are retained.

CRT1 can catalyse multiple steps in the synthesis of carotenoids. These steps require many enzymes in plants.

Recombinant plasmid

The Ti plasmid from Agrobacterium is modified using restriction enzymes and DNA ligase to delete the gall-forming gene and insert the synthesised DNA packages. A gene for resistance to the antibiotic hygromycin is also inserted so that transformed plants can be identified later. The parts of the Ti plasmid required for plant transformation are retained. Modified Ti plasmid is inserted into the bacterium. Agrobacterium is incubated with rice plant embryo. Transformed embryos are identified by their resistance to hygromycin.

Outcomes

The rice produced had endosperm with a distinctive yellow colour. Under greenhouse conditions golden rice (SGR1) contained 1.6 µg per g of carotenoids. Levels up to five times higher were produced in the field, probably due to improved growing conditions.

Recombined plasmid is inserted into Agrobacterium. This is then mixed with rice plant embryos.

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Gene for the enzyme phytoene synthase (PSY) extracted from the daffodil plant Narcissus pseudonarcissus

Gene for the enzyme carotene desaturase (CRT1) extracted from the soil bacterium Erwinia uredovora.

DNA sequences are synthesised into packages containing the CRT1 or PSY gene, terminator sequences, and endosperm specific promoters (these ensure expression of the gene only in the edible portion of the rice).

Further applications

SGR1

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Modified plants are identified by resistance to hygromycin.

Further research on the action of the PSY gene identified more efficient methods for the production of β-carotene. The second generation of golden rice now contains up to 37 µg per g of carotenoids. Golden rice was the first instance where a complete biosynthetic pathway was engineered. The procedures could be applied to other food plants to increase their nutrient levels.

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Golden rice: A controversial solution

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Golden rice (top bowl) is promoted as one way to increase the beta-carotene (and thus vitamin A) intake in countries where rice is a staple part of the diet. However there are many groups that actively resist its promotion and use (e.g. Greenpeace).

Many of these anti-GM groups believe there is no need for golden rice because people can obtain enough beta-carotene by eating a variety of fruits and vegetables. They are also worried about the possibility of golden rice contaminating ordinary rice crops.

In June 2016 the The National Academies of Sciences, Engineering, and Medicine released the results of an extensive study, reporting there was no evidence to suggest GM crops were unsafe and that GM crops are as safe to eat as non-GM crops

1. Describe the basic methodology used to create golden rice:

2. Explain how scientists ensured b-carotene was produced in the endosperm:

3. What property of Agrobacterium tumefaciens makes it an ideal vector for introducing new genes into plants?

4. (a) How could this new variety of rice reduce disease in developing countries?

5. Explain why golden rice is a controversial product:

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(b) Absorption of vitamin A requires sufficient dietary fat. Explain how this could be problematic for the targeted use of golden rice in developing countries:

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211 Engineering for Insect Resistance

Key Idea: Up to one fifth of the world's crops are lost due to insects each year. Losses can be reduced through the use of genetic engineering to introduce the Bt gene into crop plants. A key goal in horticulture is the reduction of insect crop damage. Normally this is done using sprays. However

this requires a lot of effort and leaves potentially harmful chemical residues on the food and in the environment. Using genetic engineering to produce crop plants with their own in-built insect deterrents can result in greater crop yields and reduced chemical use.

Bt toxin

In 1996 the seed company Monsanto released its first versions of Bt corn. This corn had been genetically modified to contain the gene that produces the Bt protein. The target insect pest for Bt corn is the larval stage of the European corn borer, which causes hundreds of millions of dollars worth of damage to crops annually.

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Bacillus thuringiensis is a soil living bacterium. It also occurs naturally in the gut of caterpillars and on leaf surfaces. The bacteria form spores that are associated with crystalline proteins called d-endotoxins. These are lethal to lepidopteran (butterfly and moth) larvae but do not affect other insects such as beetles or bees (or any other animal). For this reason the Bt toxin has been used as a targeted insecticide since the 1960s.

The effects of the Bt toxin on insect deterrence. The plant on the right has been treated with Bt toxin before being exposed to caterpillars. The plant on the left had not been treated with Bt toxin.

Producing a Bt plant

Genetic engineering has been used to produce cotton, corn, and potato varieties that produce the Bt toxin. The bacterium Agrobacterium tumefaciens is commonly used to transfer the Bt gene into plants, via recombinant plasmid: Agrobacterium tumefaciens

Transformed plant cells are cultured into the lab and grown into new plants before being planted out.

Bacillus thuringiensis

Agrobacterium transfers DNA into plant cell

Bt gene

Ti plasmid inserted back into Agrobacterium

Corn cell infected with Agrobacterium

Bt gene inserted into Ti plasmid

Ti plasmid

Recombinant plasmid

1. Name the bacteria that produces Bt toxin: 2. Why is Bt toxin a useful insecticide?

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4. Explain how Bt corn is produced using Agrobacterium tumefaciens:

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3. What is the primary target of the Bt toxin in Bt corn?

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What's killing the monarchs? It's not the corn

PR E O V N IE LY W So... What's killing the monarchs?

Total area occupied by monarch colonies at overwintering sites in Mexico.

22 20 18 16

Hectares (ha)

It now appears that there is a related but quite different reason for the Monarch butterfly decline. In 1996, Monsanto also began selling "Roundup Ready" corn, engineered to withstand glyphosate herbicide. Corn crops could be sprayed with herbicide and while the weeds die the corn would keep on growing, allowing less targeted spraying applications. As a result milkweed, which often grew in or near corn crops, was also killed, leaving no food for monarch caterpillars.

Above: North American populations of monarchs migrate (above) to overwintering sites in Mexico and California. Right: Monarch caterpillars feed exclusively on milkweed.

Scott McDougall

By 1999 monarch butterfly populations in the American Midwest began declining. During that year, Cornell University published a paper showing that the Bt toxin could be dispersed to other plants by the corn's pollen. Pollen landing on milkweed near corn crops could potentially kill the monarch caterpillars that fed exclusively on the milkweed. This resulted in a backlash against Bt corn by environmental activists. However, in 2001 a study was released that argued the toxin in pollen was not causing monarch decline. The toxicity in pollen was due mainly to the Bt 176 variety which was used in less than 2% of the corn grown and was in the process of being phased out. Other Bt corn varieties did not develop enough toxin, or their pollen density was too low to affect monarch caterpillars.

David R. Tribble

Bt corn was developed by the company Monsanto and sales began in 1996. There are many different types of Bt corn, each one engineered to produce the toxin in slightly different ways. One of the first produced was Bt 176.

14 12 10 8 6 4 2

Crystals of Bt toxin

1

3 -1

12

9

7

-1

10

-0

08

-0

5

06

3

1

-0

04

-0

02

-0

9

00

7

-9

98

-9

96

94

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Winter season (year)

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5. As a group, discuss the ethical issues surrounding GM corn and monarch declines. Who is to blame for the decline of monarchs and what can be done to help the population recover? Summarize the main points of your discussion below:

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212 The Ethics of Genetic Modification

Key Idea: There are many potential benefits, risks, and ethical questions in using genetically modified organisms. Genetically modified organisms (GMOs) have many potential benefits, but their use raises a number of biological and ethical concerns. Some of these include risk to human health, animal welfare issues, and environmental safety. Currently a matter of concern to consumers is the adequacy

of government regulations for the labelling of food products with GMO content. In some countries GM products must be clearly labelled, while other countries have no requirements for GM labelling. This can take away consumer choice about the types of products they buy. The use of genetic modification may also have trade implications for countries exporting and importing GMO produce.

Potential risks of GMOs

1. Increase in crop yields, including crops with more nutritional value and that store for longer.

1. Possible (uncontrollable) spread of transgenes into other species of plants, or animals.

2. Decrease in use of pesticides, herbicides and animal remedies.

2. Concerns that the release of GMOs into the environment may be irreversible.

3. Production of crops that are drought tolerant or salt tolerant.

3. Animal welfare and ethical issues: GM animals may suffer poor health and reduced life span.

4. Improvement in the health of the human population and the medicines used to achieve it.

4. GMOs may cause the emergence of pest, insect, or microbial resistance to traditional control methods.

5. Development of animal factories for the production of proteins used in manufacturing, the food industry, and health.

5. May create a monopoly and dependence of developing countries on companies who are seeking to control the world’s commercial seed supply.

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Potential benefits of GMOs

Issue: Adding genes to organisms to produce Issue: Who owns the technology? pharmaceuticals for human use or study. Background: Soybeans are the world's Background: Crop seed and animal largest agricultural crop. The United States Background: Traditionally, producing protein- breeders spend large amounts of money produces 33% of the world's soybean crop, based drugs, e.g. insulin, has been costly and on development. The genetic modification worth US$38.5 billion. Pests do a lot of unreliable. Proteins are continuously made in of plants and animals requires Government damage to the crop so reducing pest damage living organisms and these living systems can approval to develop the technology, carry out would increase crop yields and value. be exploited to produce the products need testing, and bring the product to market. This Pesticide use is common and there has been by humans. Genetic modification of plants is a lengthy process that can cost millions of a 130 fold increase in insecticide use in the and animals to produce pharmaceutical dollars. Companies therefore wish to make a US since 2001. proteins means the protein can be produced profit or at least recoup costs on their product. in large quantities relatively cheaply (once This leads to patents to protect the technology Genetic modification of crops to resist pests a stable GM organism is produced). This is and increased costs to farmers. reduces the dependence on pesticides. This already done with rennin (an enzyme used in has already been successfully performed cheese making) and human insulin (used to Problem: Biotech companies may have some in corn (Bt corn) and soybeans (resistance treat diabetes). Mammals can be genetically leverage over farmers. For example, a seed against soybean cyst nematodes). modified and induced to secrete useful producing company produces GE seeds, proteins in their milk (e.g. transgenic goats which cost a lot to develop. To ensure sales Problem: Plants that produce toxins may be produce antithrombin to prevent blood clots). and a profit, they sell only these seeds (at toxic to humans. great cost) to the farmer, who has little choice. Problem: There are concerns with animal Pests may become resistant to the pesthealth. Many genetically modified mammals GE crops may be sold to overseas markets resistant properties of the engineered plant have congenital defects and reproductive with little regulation and little choice. These rendering it ineffective. The ultimate outcome difficulties (low conception to term rates). markets will carry the load of potential of this is unknown. problems, increasing the divide between There is also the question of animal rights developed nations and developing ones. Possible solution: Careful testing of the and values. Is the genetic modification of an toxic properties of the plant under a variety of animal to produce proteins then valuing the Possible solution: Legislation must be circumstances is required to ensure it is safe protein over the animal? in place to ensure intellectual property is for human consumption. protected while also ensuring farmers have Possible solution: Testing and monitoring access to all available seed and stock. There Plans must be in place in the event that pests animal health is important. Continued must be careful consideration of the effect become resistant to the engineered plant. development of non-animal based methods of GE crops on the agriculture of developing for developing proteins. countries. LINK

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Issue: Genetically modified crops

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1. Describe an advantage and a problem with the use of plants genetically engineered to be resistant to crop pests:

(a) Advantage:

(b) Problem:

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2. Describe an advantage and a problem with using plants and animals to produce pharmaceuticals:

(a) Advantage:

(b) Problem:

3. Describe two uses of transgenic animals within the livestock industry: (a) (b)

(c) Describe the possible problems that may occur over the ownership of genetically modified organisms.

4. Some years ago, Britain banned the import of a GM, pest resistant corn variety containing marker genes for ampicillin antibiotic resistance. Suggest why the use of antibiotic-resistance genes as markers is not good practice:

5. Many agricultural applications of DNA technology make use of transgenic bacteria which infect plants and express a foreign gene. Explain one advantage of each of the following applications of genetic engineering to crop biology:

(a) Development of nitrogen-fixing Rhizobium bacteria that can colonise non-legumes such as corn and wheat:

(b) Addition of transgenic Pseudomonas fluorescens bacteria into seeds (bacterium produces a pathogen-killing toxin):

(b) Identify which of those you have listed above pose a real biological threat:

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6. Some of the public’s fears and concerns about genetically modified food stem from moral or religious convictions, while others have a biological basis and are related to the potential biological threat posed by GMOs. (a) Conduct a class discussion or debate to identify these fears and concerns, and list them below:

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213 Patterns of Disease

Key Idea: Studying the prevalence and spread of a disease gives insights into its origins and how to combat it. Diseases present at constant low levels in a population or region are known as endemic diseases. Occasionally there may be a sudden increase in the prevalence of a particular disease. On a local level this is known as an outbreak. When

an infectious disease spreads rapidly through a nation and affects large numbers of people it is called an epidemic. On rare occasions a new kind of disease will appear and spread to other countries. The rapid spread of a disease throughout the world is a pandemic. Examples of pandemic diseases include HIV/AIDS, influenza, and recently Zika virus.

Zika virus: An example of global disease spread and its containment

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Zika virus was first isolated from the area of the Zika Forest in Uganda in 1947. Since then it has spread slowly across the globe with outbreaks in the Americas in 2015 and 2016. Zika causes a mild fever and rash that is not usually serious in adults. However, in the last few years, infection of pregnant women by Zika has been linked causing microcephaly (small head and brain) in their newborns.

1954 Nigeria

2015 Cabo Verde

Movement of Zika virus across the globe

1977 Pakistan

2016 Thailand

Zika virus is carried by Aedes mosquitoes and transmitted to humans when they bite. It can also be transmitted by sexual activity, from mother to fetus during pregnancy, and in blood transfusions.

2007 Yap

1947 Uganda

1948 1975/2010 Tanzania Gabon

2016 Venezuela

2016 Fiji

2016 1977/2012 Maldives Malaysia

2013 Cook Islands

The severe effects of Zika on fetal development prompted world health authorities to begin an awareness campaign to limit Zika's spread and reduce the risk of contracting it. The campaign has focussed on prevention, including travel advisories in unaffected countries and awareness campaigns in affected countries.

Insect repellent should especially be used if wearing clothing that exposes the skin.

2015 Brazil

2014 Easter Island

Zika virus became an important international concern in 2015 and 2016 in the lead up to and during the 2016 Rio Olympics. Concerns focussed on the movement of spectators, tourists, and athletes and the spread of the disease around the globe as they returned home after the events.

Reducing areas where water can stagnate reduces mosquito breeding sites.

People are advised to wear long sleeves and pants to prevent mosquito bites.

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1. (a) Which general direction has Zika virus spread across the globe?

2. How is Zika virus transmitted?

3. Describe how the spread of Zika virus can be reduced:

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(b) Describe the area that Zika virus appears to be generally confined to and explain this:

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214 Containing the Spread of Disease risk that diseases of humans, livestock, and crops will be spread between countries. Australia is fortunate in that its geographical isolation has helped to prevent the spread of disease from other parts of the world. Precautions such as quarantine, which isolates exposed individuals that may be infected, as well as screening of imported produce and international travellers, help to limit the entry of diseases into Australia. Quarantine is distinct from isolation, which aims to contain disease by isolating an already infected person.

Transmission of disease can be reduced by adopting 'safe' behaviours. Examples include using condoms to reduce the spread of STIs, isolation of people already infected, or establishing quarantine procedures for people who have been exposed to infection.

The environment can be made less suitable for the growth and transmission of pathogens. For example, spraying drainage ditches and draining swamps eliminates breeding habitats for mosquitoes carrying diseases such as malaria and dengue fever.

Disinfectants and sterilisation techniques, such as autoclaving, destroy pathogenic microbes before they have the opportunity to infect. The use of these techniques in medicine has significantly reduced post operative infections and associated deaths.

The development of effective sanitation, sewage treatment, and treatment of drinking water has virtually eliminated dangerous waterborne diseases from developed countries. These practices disrupt the normal infection cycle of pathogens such as cholera and giardia.

Appropriate personal hygiene practices reduce the risk of infection and transmission. Soap may not destroy the pathogens but washing will dilute and remove them from the skin. Although popular, antibacterial soaps encourages the development of strains resistant to antimicrobial drugs.

Vaccination schedules form part of public health programmes. Vaccination is one of the most effective ways of preventing transmission of contagious diseases. If most of the population is immunised, herd immunity limits outbreaks to sporadic cases and prevents epidemics.

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Key Idea: Preventing the entry and spread of pathogens is important in protecting a country's population and industries from infectious diseases. Many factors can influence the spread of disease, including the social climate, diet, general health, and access to medical care. Human intervention and modification of behaviour, including vaccination, can reduce the transmission rate of some diseases and inhibit their spread. Global air travel and international trade in commodities has increased the

(b) How is disease transmission reduced in medical care situations?

(c) Why is sanitation important in preventing the spread of disease?

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1. (a) Identify three ways in which the environment can be made less suitable for establishment and transmission of diseases:

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The disease risk

The Australian Quarantine Inspection Service (AQIS) plays an important role preventing the entry of exotic pests and diseases that could affect plant, animal, and human health, and the environment. All passengers and freight entering Australia by air or sea (even small, private vessels) are required to pass inspection by a Quarantine Officer. Australia has strict quarantine regulations because introduced animals and plants could also bring in pests and diseases that have no natural enemies or controls.

Communicable diseases spread from one host to another, either directly or indirectly. Those that are easily spread, such as measles, are said to be contagious. Such diseases are a threat to public health and many must be notified to health authorities. Many of the serious contagious bacterial and viral diseases of the past are now controlled through vaccination programmes. Even if susceptible individuals contract the disease, the spread is limited because of the herd immunity of the generally vaccinated population.

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Goods PROHIBITED entry into Australia without authorisation • Milk and dairy products • Seeds and beans • Popping corn and raw nuts • Eggs and egg products • Fresh fruit and vegetables • Live animals • Meat and fish products • Live plants • Biological materials (e.g. human/animal vaccines) • Deer horn/velvet • Soil and sand

Noncommunicable diseases, such as chronic lung diseases, are not spread from one host to another and pose less threat to public health. A disease that occurs only occasionally and is restricted in its spread is called a sporadic disease (e.g. Legionnaire's disease).

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The role of quarantine

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English measles (top) and whooping cough are both controlled by vaccination

2. (a) Explain why contagious diseases present such a threat to public health:

(b) How are contagious diseases generally best controlled?

(c) Why is this method of control effective in limiting the spread of the disease?

3. Distinguish between communicable disease and noncommunicable diseases:

4. (a) How does quarantine prevent the spread of disease?

(b) Distinguish between quarantine and isolation and describe the role of each in containing the spread of disease:

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5. How does Australia's geographic position help prevent the entry and spread of disease in Australia?


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215 Emerging Disease malaria) has lead to the re-emergence of diseases that were previously thought to be largely under control. In the 1940s, many common but lethal bacterial diseases (e.g. diphtheria) were conquered using antibiotics. It is now evident that antibiotics are losing efficacy and must be used wisely to reduce the risk of resistance in the target populations. The global spread of viral diseases is controlled most effectively by immunising susceptible populations against circulating strains (as occurs with seasonal flu). The challenge for the control of viral disease is in the continual development of effective vaccines against newly emerging strains.

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Key Idea: Emerging diseases are diseases that are new or have increased in incidence in the human population in recent times and as such pose a risk of rapidly spreading. Emerging diseases are so named because they are new to humans or have increased markedly in incidence in recent times. Often, as with HIV/AIDS and avian influenza (H5N1), they are zoonoses (animal diseases that cross to humans). Zoonoses are capable of causing highly lethal pandemics (world-wide epidemics) amongst an unprepared population. The increasing incidence of multiple drug resistance in pathogens (including those that cause tuberculosis and

Many emerging diseases originate in animals and cross to humans (zoonoses). As humans put more pressure on wild spaces they come into contact with animal populations that may be the reservoirs (long-term natural hosts) of diseases new to humans. Some of these diseases are capable of not only crossing the species barrier to humans, but also being transmitted between humans.

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Coronaviruses

1. (a) Where do many emerging diseases come from?

Hu

(b) Suggest why diseases of these origins are becoming more prevalent:

S

Coronaviruses cause respiratory infections. The SARS (Severe Acute Respiratory Syndrome) outbreak in 2003 caused over 8,000 cases and 774 deaths in 34 countries.

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MERS (Middle East Respiratory Syndrome) was first recorded in 2012. About a 1000 cases have been reported with 40% fatality.

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Transmission host (usually moderate symptoms)

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Spillover host (severe infection and usually high mortality)

Henipaviruses are found naturally in fruit bats. In the case of Hendravirus, horses are infected by bats. Humans may be infected from infected horses. Fatality of Hendravirus cases in horses is 75% and in humans 60%.

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Influenza is a common virus. However variations of this virus originating from birds have swept the world in the last few decades. The fear is that influenzas in birds and pigs could jump to humans causing major flu pandemics (e.g. the swine flu pandemic of 2009).

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2. Why are new diseases that can be transmitted between humans of such global significance?

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216 Identifying and Treating Disease

Key Idea: Diagnostic techniques and treatments for disease are becoming more advanced as scientific knowledge advances and the tools available become more sophisticated. The correct diagnosis of a disease is extremely important for its correct treatment. Traditional methods of diagnosing

a disease include observing patient symptoms (e.g. type of rash, origin of any pain etc) and observing the morphology of the pathogen and its biochemical reactions. Newer methods of identifying a disease use molecular techniques (including DNA profiling) to pinpoint the type of pathogen.

Identifying a pathogen

Immunology

Morphology (bacteria and fungi)

ff Agglutination tests. An antibody is

ff Cell shape (spherical, rod shaped, or spiral shaped).

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mixed with patient serum. If the antigen (pathogen) is present, crosslinks form between the particles. Agglutination tests are not sensitive in that they cannot identify the pathogen but are useful for establishing presence or absence.

ff Gram stains are simple to perform and identify a bacterial pathogen as either gram negative or gram positive. These classes respond differently to antibiotics.

ff Special structures (e.g capsule) ff Bacteria can be grown on agar plates.

ff Precipitation tests measure the degree of precipitation in serum and again are useful for presence/absence testing.

ff A Western blot test detects antibodies by the reaction between them and antigens immobilised on a gel. Western blots generally have good sensitivity although less than ELISA, which is highly specific.

ff ELISA test (below) - enzyme linked

Genetic methods

ff Similarly, different types of fungi

ff DNA profiling enables the identification

will produce identifiable colonies of characteristic shapes and colours.

of pathogens that are difficult to culture to the lab (e,g. viruses).

ff Adding different nutrients to the agar

ff PCR isolates sections of DNA that can

plate affects the bacteria that can grow, allowing identification. Once bacteria are growing, different antibiotics can be added to discover the best treatment.

be used to produce a profile. This can be compared to known profiles to identify pathogens such as viruses.

Rkalendar CC 3.0

immunosorbent assay. Antibodies with attached markers are used to attach to and identify the pathogen's antigens.

The shape of the colony on the plate can be used to identify the type of bacteria.

1. Why is correct diagnosis of a disease and its pathogen important?

2. Why would morphology testing be appropriate for fungal and bacterial pathogens?

3. (a) Which method(s) listed above are more appropriate for identifying viral infection?

4. Explain what a test's sensitivity means and why it is important:

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(b) Explain why:

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For most known and common pathogens a treatment programme is already established and can be implemented quickly once the pathogen is identified. Sometimes further testing is needed to identify treatment for new or particularly virulent strains, or when the extent of the infection is not know. Considerations before treatment include:

Timing of antimicrobial treatment Urgent cases may be treated with broad spectrum antibiotics, but less urgent cases should wait until diagnosis is complete. Prematurely starting treatment may result in inappropriate treatment or may even affect diagnosis from subsequent blood samples.

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Treating one pathogen may require several drugs Using a combination of drugs is useful when drugs have a synergistic effect (their combined effect is greater than the sum of the two individual drugs). It is also used when a broad spectrum approach is required before targeted treatment can begin.

Uwe Gille cc 3.0

Michael Berry cc 3.0

Bactericidal vs bacteriostatic therapy Bacteriostatic drugs inhibit bacterial growth. Bactericidal (agents that kill the bacteria) are preferred in acute and serious cases of infection.

Results of susceptibility tests Once a pathogen is identified, antimicrobial susceptibility testing can determine the ability of the pathogen to grow in the presence of a particular drug. Clearance around the antibiotic disc (above) indicates susceptibility to that drug.

Oral or intravenous therapy. Patients with severe infections are often given antimicrobial drugs intravenously as the drugs are available immediately and are more easily utilised by the body. IV also enables simultaneous delivery of multiple drugs (useful if patients are very ill).

Using antimicrobial combinations Combining drugs is important when combating more than one pathogen and in reducing the risk of the pathogens developing resistance during long term treatments. Combination therapy is used in the treatment of tuberculosis, leprosy, and HIV/AIDS.

(a) Use of a combination of drugs:

(b) Bactericidal vs bacteriostatic therapy:

(c) Oral or intravenous therapy:

(d) Timing of treatment:

(e) Results of susceptibility test:

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5. Explain the importance of each of the following in treating a disease:

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217 Rational Drug Design

Key Idea: Analysis of genomic information could lead to new and more efficient ways of treating or preventing disease. Rational drug design is the process of finding and producing new medications based on the biology of the target organism.

The drug will have a specific target, e.g. an enzyme in a metabolic pathway or the antigenic portion of a pathogen. Using databases, potential target molecules can be identified and drugs designed on the basis of the target's structure.

Rational drug design is based on two main ideas.

ff That a target for therapy can be modulated, i.e.

a change to it will have some therapeutic effect.

ff The target is 'druggable', i.e. it is capable of

Potential drug can bind to the active site and decrease enzyme's activity.

Potential drug

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binding a molecule and its activity can be changed by binding the molecule.

Target molecule (enzyme)

3D structure allows visualisation of active area or receptor.

ff Once a potential drug is discovered it

becomes a lead compound. A lead compound is a compound on which to base further drug design.

ff Molecules related to the lead compound

are screened by high throughput screening. Thousands of compounds can be tested at the same time and their biological activity determined. Molecules with the best activity can then be targeted for further testing. Molecules are tested for the activity and potency of the compound before being sent for development and clinical testing.

ff A target for therapy can be visualised by using

3D computer software and this may provide clues to the shape and charge of the molecule (drug) needed to bind to it. Databases can be used to scan for molecules that may be useful.

One of the new directions of research in medicine is the development of drugs targeting the gene function or gene products (proteins) of a pathogen. This begins with sequencing a genome and adding it to the database of already sequenced genomes. Genes are identified, cross referenced with other known similar genes, and their functions investigated. Any existing drugs targeting similar genes or gene products can then be identified and their effectiveness against the newly identified genes and their products tested. Novel genes can be identified and ways of exploiting them for medical purposes can be investigated.

Malaria is a disease caused by the protozoan parasite Plasmodium of which P. falciparum is the most deadly. Plasmodium is becoming increasingly drug-resistant so a vaccine offers the best hope of controlling the disease.

1. Explain how the use of computers is aiding drug design:

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Plasmodium sporozoite

The genome of P. falciparum was published in 2002. Fifteen loci have been identified as encoding antigens that may be useful in vaccines, including an antigen-rich region on chromosome 10. However only six of the loci appear to be similar to other Plasmodium species, reducing the likelihood of developing a single vaccine effective against all species.

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Once the gene is identified, the gene product (protein) is synthesised in the lab and tested in a model organism for an immune response. If successful, the product can then be used as the basis of a vaccine.

Ute Frevert

Reverse vaccinology is a technique in which the entire genome of a pathogen is screened for genes that may produce antigenic properties, e.g. genes that code for extracellular products such as surface proteins.

National Human Genome Research Institute

Reverse vaccinology and malaria

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Many of the medicinal drugs we use are based on plants and plant extracts. Plant compounds are constantly being investigated and tested and for their antimicrobial properties. Below are examples of plant compounds known to have antimicrobial properties.

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Rosemary

The antimicrobial compound in chili peppers, is capsaicin, an antibacterial agent of relatively high potency.

The essential oil produced from the herb rosemary acts as an effective general antimicrobial agent.

Phloretin is a polyphenol extracted from the leaves of apple trees. It acts as a highly potent antimicrobial.

Saponins (compounds that foam in water) extracted from ginseng are effective against a range of bacteria.

Isolating and testing antimicrobial compounds from plants

High antimicrobial activity

Solvent extraction

Pulping

Extracted compound

No antimicrobial activity

A plant is chosen for testing. Plants produce a wide range of compounds, but those with aromatic rings often have medicinal properties.

Extraction: Plant material is pulped, then soaked in a solvent (water, alcohol), or boiled to release and crudely fractionate the compounds.

Screening: Microbes are plated onto agar. Paper discs soaked with plant extract are placed on agar, or the plant extract is pipetted into wells cut into the agar. The plates are then incubated. The effectiveness of the extract can be determined by measuring the clear zone around the plant extract after incubation.

Separation and analysis: Mass spectroscopy is used to further separate and isolate compounds from the extracts that show antimicrobial activity.

Concentration: The compounds may be further separated, (e.g. by chromatography), before they are concentrated using freeze drying or rotary evaporation (above).

Testing: The isolated and concentrated compounds are tested for antimicrobial activity. This identifies which component of the initial extract is the antimicrobial.

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3. Why is research into antimicrobial plant extracts increasing?

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2. Explain how the completion of P. falciparum genome has helped make the development of a malaria vaccine more likely:

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218 Relenza: How an Antiviral Drug Works

Key Idea: Relenza is an antiviral drug designed to inhibit the spread of influenza. Relenza was designed by rational drug design in 1989 by the Australian CSIRO. The drug was designed to target the

enzyme neuraminidase, which is needed by the influenza virus to leave a host cell. By inhibiting neuraminidase activity, the influenza virus becomes trapped in the host cell and is unable to spread.

Mechanism of operation

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The influenza virus is surrounded by H (haemagglutinin) and N (neuraminidase) spikes. The different variations in these spikes give the H-N- notation of influenza viruses (e.g. H1N1). X-ray crystallography of viral particles revealed that a region on the neuraminidase protein that cleaved the link between the virus and host cell remained constant between strains and could be targeted by an inhibitor. Haemagglutinin spike

Cell infected with influenza virus

Neuraminidase triggers virus release by cleaving link between cell and virus.

Relenza binds to neuraminidase active site.

Receptor

Cell infected with influenza virus

Viruses free to infect other cells.

Virus unable to leave infected cell. Infected cell is destroyed by immune system.

Neuraminidase spike

Neuraminidase inhibitors just a placebo?

Outbreaks of influenzas, such as swine flu (right), over the last few years have shown why drugs that are effective against flu are important. Health organisations such as WHO have urged governments to stockpile flu drugs such as Tamiflu and Relenza in case of major outbreaks. However data released in 2014 and based on the original trials of some 24,000 people have suggested these drugs have little or no effect on the flu. Results show that, on average, a person taking these drugs will have flu symptoms for 6.3 days. Flu symptoms normally last about 7 days. In addition the drugs caused a 4-5% increase in nausea in adults and a 1% increase in the risk of a psychiatric event (e.g. delirium or hallucinations).

1. (a) What is the role of the haemagglutinin spike on an influenza virus?

(b) What is the role of the neuraminidase spike on an influenza virus?

(b) What feature of the spike made it possible to target all strains with an inhibitor?

3. Explain how Relenza inhibits the spread of the influenza virus:

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2. (a) How was the neuraminidase spike studied?

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219 Antibiotics Vs Antivirals selective toxicity, killing the pathogen without damaging the host. The best antibiotics have a narrow spectrum of activity and target only the bacterial species responsible for the infection. Broad spectrum antibiotics target a large number of bacterial species and so affect the host's natural microflora as well as the pathogen. While antibiotics are effective against bacteria, they do not affect viruses, which are not prokaryotes and have no metabolism of their own. Antiviral drugs are used to inhibit various stages of the viral life cycle, from penetrating the host cell to disrupting the release of the virus.

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Key Idea: Antibiotics work against bacteria but are ineffective against viral pathogens, which require special antiviral drugs. An antibiotic is a substance that inhibits or prevents bacterial growth. Antibiotics are produced naturally by bacteria and fungi against their competitors and this ability has been exploited by humans for the control of bacterial infections. Antibiotics interfere with bacterial growth by interfering with key aspects of bacterial metabolism (below). Their action may kill the bacterial cells directly (bactericidal) or inhibit their growth (bacteriostatic). The ideal antimicrobial drug has

How antibiotics work

Rupture plasma membrane Some antibiotics disrupt the plasma membrane by interacting with the phospholipids. Such antibiotics are often used as a last resort because they are also quite toxic to the host.

DNA

Inhibit protein synthesis The enzyme responsible for elongating the polypeptide chain is inhibited so protein synthesis stops.

Inhibit cell wall synthesis Some antibiotics stop the synthesis of new cell walls during cell division.

Translation

Transcription

Protein

Replication

mRNA

Inhibit gene copying Some antibiotics prevent DNA replication and transcription

Inhibit enzyme activity Some antibiotics inhibit the synthesis of essential metabolites.

1. (a) What are the requirements of an "ideal" antimicrobial drug?

(b) In what way do antibiotics satisfy these requirements?

2. Why would antibiotics that target the plasma membrane also be harmful to the host?

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4. Explain why antibiotics do not work on viruses:

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3. Explain the difference between bacteriostatic and bactericidal antibiotics:

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325 Mechanisms of antiviral drugs

Currently, antiviral drugs are available for use against HIV, hepatitis, influenza, and the herpes virus. Most antiviral drugs work by inhibiting replication of the virus. The drug stops the virus replicating giving the immune system time to destroy the virus. Antiviral drugs work on several parts of viral life cycle (right).

Block receptor binding

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Block virus entry into cell

Inhibit uncoating

Inhibit integration, nucleic acid synthesis, transcription, and translation.

Oseltamivir (marketed as Tamiflu, above) slows the spread of influenza A and B viruses in the body by preventing a newly formed virus particle from budding from an infected cell. However, various trials have shed doubt of whether the drug produces any real benefit to flu sufferers.

Inhibit assembly

Block exit from cell

5. Two students investigated the effect of antibiotics on bacterial growth. They placed antibiotic discs on petri dishes evenly coated with bacterial colonies. Dish 1 contained four different antibiotics (A-D) and a control (CL). Dish 2 contained four different concentrations of a single antibiotic and a control (CL).

Zone of inhibition (no bacterial growth)

3

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Paper discs saturated with antibiotic

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7 µg mL-1

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5.5 µg mL-1

4

Bacterial colonies (bacterial lawn)

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Plate 1

Plate 2

(a) Which was the most effective antibiotic on plate 1? Explain your choice:

(b) Which was the most effective concentration on plate 2? Explain your choice:

(a) Blocking receptor binding:

(b) Inhibiting assembly:

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6. For each of the following mechanisms used by antiviral drugs explain how it prevents spread of the virus:


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220 KEY TERMS AND IDEAS: Did You Get It?

1. Test your vocabulary by matching each term to its correct definition, as identified by its preceding letter code.

A An organism or artificial vehicle that is capable of transferring a DNA sequence to another organism.

forensics

B The manipulation of DNA and gene sequences in order to modify the characteristics of organisms.

gene technology

C The process of locating regions of a DNA sequence that are variable between individuals in order to distinguish between individuals.

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DNA profiling

genetic modification

D A small circular piece of DNA commonly found in bacteria.

GMO

E DNA that has had a new sequence added so that the original sequence has been changed.

marker gene

F A short (normally two base pairs) piece of DNA that repeats a variable number of times between people and so can be used to distinguish between individuals.

microsatellite

molecular clone

G A gene, with an identifiable effect, used to determine if a piece of DNA has been successfully inserted into the host organism.

H One of multiple copies of an isolated defined sequence of (often recombinant) DNA.

plasmid

I

recombinant DNA

J The branch of biotechnology that involves manipulating the genetic constitution of organisms in some way.

transgenic organism

K The application of scientific methods to the investigation of matters involving criminal and civil laws.

vector

An organism that has had part of its DNA sequence altered either by the removal or insertion of a piece of DNA.

L An organism that has had its genome modified to carry genes that are not usually found in that species.

2. The electrophoresis gel (below, right) shows four profiles containing five STR sites: the mother (A) her daughter (B) and two possible fathers (C and D). Which of the possible fathers is the biological father?

(a) The biological father is:

(b) Why do profiles B and D only have 9 bands?

A

B

C

D

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3. Distinguish between antiviral drugs and antibiotics and explain the significance of the difference:

TEST

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4. Distinguish between VNTRs and STRs and outline their uses in modern genetics:

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221 Review: Unit 4, Area of Study 2

Summarise what you know about this topic under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts. Use the images and hints to help you and refer back to the introduction to check the points covered:

Biological knowledge and society

HINT: Principles and uses of DNA profiling, production and use of transgenic organisms. Identifying, treating, and containing disease.

DNA manipulation

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HINT: Making and using recombinant DNA, PCR and gel electrophoresis.

REVISE


222 Synoptic Assessment: Unit 4, Area of Study 2

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In this activity you will use your knowledge of genetic engineering and biotechnology to genetically engineer a food crop to maximise its growth under adverse conditions.

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Currently 1/6 of the world's population are undernourished. If trends continue, 1.5 billion people will be at risk of starvation by 2050 and, by 2100 (if global warming is taken into account), nearly half the world's population could be threatened with food shortages. The solution to the problem of food production is complicated. Most of the Earth's arable land has already been developed and currently uses 37% of the Earth's land area, leaving little room to grow more crops or farm more animals. Development of new fast growing and high yield crops appears to be part of the solution, but many crops can only be grown under a narrow range of conditions or are susceptible to disease. Moreover, the farming and irrigation of some areas is difficult, costly, and can be environmentally damaging. Genetic modification of plants may help to solve some of these looming problems by producing plants that will require less intensive culture or that will grow in areas previously considered not arable. Your task is to use the items shown to devise a technique to successfully create a plant that could be successfully farmed in semi-desert environments such as sub-Saharan Africa. The following page will take you through the procedure. Not all the items will need to be used.

Enzymes

Useful organisms

Restriction enzyme

Fungus that is able to survive dry conditions using two enzymes WA-UT1 and Ter-UT2 to facilitate water uptake.

Reverse transcriptase

Bacterium known to thrive in dry conditions using a single enzyme DRI-X1 to catalyse multiple reactions.

DNA ligase

Plant identified for modification

Petri dish

Adenovirus

Retrovirus

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Plasmid

Agrobacterium

Equipment

TEST

Liposome

Possible Vectors

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Incubator

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1. Identify the organism you would chose as a 'donor' of drought survival genes and explain your choice:

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2. Describe a process to identify and isolate the required gene(s) and identify the tools to be used:

3. Identify a vector for the transfer of the isolated gene(s) into the crop plant and explain your decision:

4. Explain how the isolated gene(s) would be integrated into the vector's genome:

5. (a) Explain how the vector will transform the identified plant:

(b) Identify the stage of development at which the plant would most easily be transformed. Explain your choice:

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7. Identify any ethical issues associated with this technique:

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6. Explain how the transformed plants could be identified:


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Practical investigation

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Unit 4 Outcome 3

Key terms

Experimental design

accuracy

Key knowledge and skills

aim

TASK: Design or adapt a lab or field-based investigation related to cellular processes and/or biological change and continuity over time. You are required to identify an aim, develop a question, formulate a hypothesis, and plan a course of action to answer the question, complying with safety and ethical guidelines.

assumption control

controlled variable

dependent variable ethics

fair test

1

Show an understanding of the biological concepts and key terms as relevant to your investigation.

225 226

c

2

Demonstrate an understanding of the characteristics of scientific research methods and techniques for collecting primary data as relevant to your investigation, be it laboratory work (cytology, biochemistry, immunology) or fieldwork (geomorphology). Determine your aim and hypothesis, ask questions, and make predictions that can be tested.

226 227

c

3

Identify dependent, independent, and controlled variables. Explain your choice of a control, and show awareness of assumptions in your investigation.

225 226

c

4

Show understanding of precision, accuracy, reliability, and validity when collecting your data. Precision, accuracy, and reliability are features of the assessment or measurement tools used, whereas validity is a feature of design.

225 226

independent variable precision

prediction

primary data

quantitative data

Precision: How close your measurements are to each other.

qualitative data

Accuracy: How close your measurements are to the true value of the variable.

reliability

Reliability: How much your assessment tool provides consistent, stable results.

sample

Validity: The extent to which the study measures what it is intended to measure.

scientific poster

Explain how you will minimise bias, ensure accuracy (e.g. through calibration of equipment, and maximise precision of measurements.

validity (of data)

c

5

Demonstrate an ability to carry out an investigation safely and ethically.

223

PASCO

Analysis and presentation of research Key knowledge and skills

Activity number

224 225

6

Use appropriate means to organise, analyse, and evaluate primary data to identify patterns and relationships. Include reference to sources of error and limitations of data or methods.

c

7

Understand how models, theories, and classification keys are used to organise and explain observations and biological concepts and recognise their limitations.

226

c

8

Make clear, accurate scientific drawings as appropriate to your investigation.

226

c

9

Present and explain the key findings of your investigation as a scientific poster presentation to include a title, introduction, methodology, results, discussion, conclusion, and references and acknowledgements. Include appropriate biological terminology and representations, standard abbreviations, and units of measurement, and acknowledge all sources of information.

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c

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variable

226

c

hypothesis

observation

Activity number


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223 Ethics and Safety Issues of Research

Key Idea: When designing an investigation all potential health and safety risks must be first identified and steps taken in the design process to reduce or eliminate them. When designing an experiment it is important to identify any potential hazards to health and safety and take steps to eliminate or reduce the risks they pose. A failure to do

this could result in harm to yourself or others around you. Ethical considerations also need to be addressed. These include bioethics (the moral implications of new discoveries), animal welfare issues, and your own behaviour in recording reporting results accurately and honestly and acknowledging the work of others.

Heath and safety considerations

Reducing heath and safety risks

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Laboratory hazards fall into three general categories: chemical, biological, and physical. Depending on the hazard, they have potential to cause harm to people, other organisms, or the environment.

►► Chemical: Chemicals could be ingested, absorbed through the skin, or inhaled. Examples include cleaning agents, disinfectants, and reagents (powdered and liquid). Some chemicals can cause fires or explosions if not handled correctly. ►► Biological: All biological material should be treated as potentially hazardous to avoid contamination and possible harm. Examples include microbial samples, animal tissue, fluid samples, and plant samples.

►► Identify potential hazards before you start and become knowledgeable about their risks. ►► Wear appropriate safety gear (lab coat, gloves, safety glasses, ear protection, and a mask as necessary).

►► Physical: There are numerous potential physical hazards ranging from the laboratory environment itself to the equipment you are using. Common hazards include injury caused by not using the equipment correctly (electrical, thermal, or sound hazards), cluttered working spaces, and tripping or slip hazards (e.g. wet floor).

►► Ensure all chemicals and solutions are clearly labelled. ►► Know how to correctly use all equipment and machinery before you begin. ►► Maintain clean work spaces and floors to reduce the risk of slips and spills. Keep access ways to emergency equipment clear.

Report your true data and findings, even if they are not the results you were expecting. Changing results to fit your hypothesis is misleading and unethical.

It is very important to acknowledge the work of others (e.g. photographs, data, reference material). Failure to do so is plagiarism.

Janet Stephens

Ethical considerations

Get a teacher to review your experimental design for ethical approval prior to beginning. Minimise the impact of your research on the environment.

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(b) What has been done to reduce one of these risks?

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1. (a) Identify potential health and safety issues associated with the dissection of the pig kidney being carried out in the photo (left):

KNOW


224 Maintaining a Log Book

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332

Key Idea: A logbook records your ideas and results throughout your scientific investigation. It provides a record of work and proof that you have carried out that work. A logbook provides a complete record of the ideas and work you have carried out during your investigation. Each entry

Find a notebook to use that will suit your purposes (e.g. if the logbook is to be used in a field a waterproof book is useful). A hardback A4 lined exercise book is a good choice, anything smaller will make it difficult to include photos or extra pages later on.

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1

must be dated and it should show in detail how you carried out your experiment or research project. The logbook provides proof that you have carried out certain activities on specific dates, and records the results of your work. Logbooks can be used to verify the authenticity and originality of your ideas.

2

Name your logbook in a prominent location. Number the pages so you can create a good table of contents. Creating sections in your logbook with tabs helps you keep track of ideas, methods, and results easily.

3

Date and sign every entry. Entries should be concise, but contain enough information that you can understand them later on. Short notes and bullet points are often used.

4

You must be able to read your entries at a later date, but don't worry too much about neatness. Logbooks are a record of your work, not the final report. It is more important to accurately record information during lab trials or field studies than to have a nice looking logbook!

5

Your logbook should be used in all phases of your investigation, from planing to write up. Record ideas on methods or analysis, as well as results.

Date your entries

7

Include any mishaps, failed experiments, or changes in methodology in your logbooks. Where possible, explain the reasons for the failure or change. Sometimes failed experiments can be just as valuable as successful experiments in understanding a result.

8

Include all observations made during your investigation and any calculations and transformations of the data.

9

Remember, systematically recording your ideas, observations, and analyses during your investigation will pay off when you have to organise the material for the final write up. It will also help to clarify any parts of your study that your teacher or marker may find confusing or incorrect, meaning you could still get credit for your work.

1. Why is it important to keep a detailed logbook during a scientific investigation?

KNOW

Calculations should be included

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Glue, staple, or tape any loose paper or photos into your logbook. Loose papers are an annoyance, both for you to keep track of and your teacher to sort through.

Staple or fix loose paper and photos

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6

Include sketches and ideas

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225 Analysis and Interpretation

Key Idea: Once you have collected your data you must select an appropriate form of analysis to establish if your hypothesis is supported and identify any trends or patterns in the data. Data analysis involves examining and processing the data you have collected to identify trends and patterns and establish whether or not the data support your hypothesis and help to answer the questions you posed in your investigation.

23%

Water use key

17%

Cooling water

There are many ways to analyse and present data and your Irrigation choice must be appropriate for the data you 27%have collected. Commercial 33% A statistical analysis is sometimes necessary but for simple, /washwater well designed experiments basic descriptive statistics (e.g. Drinking supply means) may be all that is needed. Finally you must choose the best way to display your results (e.g. table or graph) so that you can present your data in an organised way. Average household water consumption in Australian cities How do I analyse my data?

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Review your initial data

Sydney

►► After you have collected your first set of data (or your preliminary data) it is a good idea to spend a 23% short period of time analysing it.

17%

►► You may discover that you need to collect your data differently to how you first planned (e.g. taking more 33% measurements or changing the way you collect your data, such as automation for rapidly occurring changes or prolonged time series data).

►► Check your data to see that it makes sense. Do the results seem logical? Are there any outliers? If so, you must Perth Water use key decided whether to include them in your analysis. Cooling water

Melbourne

►► Raw data may need to be transformed to see trends Irrigation Adelaide 27% and patterns. Recall that these transformations are Commercial Canberra rates, ratios). Other often quite simple (e.g. percentages, /washwater transformations are used to normalise the data so that it Drinking supply Hobart can undergo further analysis (e.g. log transformations when 0 200 400 600 working with large numbers).

►► Take some time to plot the data or calculate Household consumption (L per year x 1000) summary statistics as these will allow you to see ►► Descriptive statistics (e.g. mean and standard deviation) trends and patterns more easily than whenAverage the datahousehold water provide a way to summarise your data, and provide results consumption in Australian cities is recorded in a log book. Once you are satisfied that can easily be presented and compared across groups. that your methods of data collection are adequate Summary statistics are also useful in identifying trends and Sydney you can continue with your investigation. patterns in the data. Melbourne Adelaide

Canberra

►► Sometimes an appropriate statistical analysis is required to test the significance of results. However, with simple experiments, if the design is sound, the results are often clearly shown in a plot of the data. Frequency

Perth

Hobart

Weight (g)

0 200 400 600 Household consumption (L per year x 1000)

14

Temperature vs metabolic rate in a rat Line connecting points

12

17%

Water use key

Cooling water

33%

27%

Irrigation

Metabolic rate

23%

Frequency

10